5 Integrals reviewed Basic facts U-substitution... 4

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1 Contents 5 Integrals reviewed 5. Basic facts U-substitution Integral Applications 0 6. Area between two curves Volumes by rotation Volumes by Cylindrical Shells Average Value of a Function Techniques of Integration Integration by Parts Trigonometric Integrals Trigonometric Substitution Integration of Rational Functions Strategy for Integration Improper Integrals Further Integral Applications 7 8. Arc Length Probability Sequences and Series 8. Sequences Series The Integral Test The Comparison Tests Alternating Series Absolute Convergence and the Ratio and Root Tests Strategy for Series Power Series Taylor and Maclaurin Series Parametric and Polar Polar Coordinates Curves Defined by Parametric Equations Calculus with Parametric Curves Areas in Polar Coordinates

2 CONTENTS A Absolute values 46 B Trig functions 47 C Horizontal Asymptotes 50 D Extended Reals 5 E L Hospital s Rule 5 F Cartesian and Polar Graph paper 55

3 Chapter 5 Integrals reviewed 5. Basic facts Definition. Let f(x) be a function defined on the interval [a, b].,, 3,... we define a Riemann Sum (with n steps) as If n equals f(x ) x + f(x ) x + + f(x n ) x. We assume that x, x,..., x n are equally spaced x-values in the interval [a, b], each a distance of x from the previous one, where x = b a n. Definition. Let f(x) be a function defined on the interval [a, b]. Suppose that f(x) 0 for all x in the interval. Let A be the area between the graph of f(x) and the x-axis, and between x = a and x = b as shown f(x) A a b Then we can find A using a Riemann Sum as shown A = lim n (f(x ) x + f(x ) x + + f(x n ) x). Comments. To understand this fact, re-interpret each term f(x i ) x as an area f(x i ) x = Area of a rectangle f(x i) For instance, we can re-interpret f(x ) x + f(x ) x + f(x 3 ) x + f(x 4 ) x as the area of four rectangles added together, as shown in the picture 3 x

4 CHAPTER 5. INTEGRALS REVIEWED 4 f(x) x x x 3 x 4 In a similar way, we can re-interpret the Riemann Sum with a large value of n as the sum of the area of a large number of rectangles: 5 rectangles Definition. Let f(x) be a function defined on the interval [a, b]. We define the symbol b a f(x) dx as follows b a Interpretation If f(x) 0, then f(x) dx = lim n (f(x ) x + f(x ) x + + f(x n ) x) b a f(x) dx can be interpreted in terms of area: it is the area between the graph of f(x) and the x-axis, and between x = a and x = b. In general, b a f(x) dx can be interpreted as signed area: this means that any area below the x-axis is counted as negative. If f(x) is made of simple shapes such as rectangles, straight lines, parts circles, etc., then b with negative signs). In other applications a f(x) dx can be found by using basic formulas for area (possibly b a f(x) dx can be interpreted as: area between curves, volumes of revolution, arc-length, surface area, probability, etc. All of these can be understood by looking at Riemann Sums: whatever f(x i ) x repre- If part of the graph is above the x-axis and part below, then we can split the whole integral up into pieces, each piece being entirely on one side of the x-axis. Then the whole integral is the sum of the integrals of the pieces. When we add the integrals for pieces below the axis, we are adding negative areas, i.e. subtracting.

5 CHAPTER 5. INTEGRALS REVIEWED 5 sents, we can understand (and calculate an approximation of) a sum of quantities of the same kind as f(x i ) x. Theorem (Fundamental Theorem of Calculus). b where F (x) is any anti-derivative of f(x). a f(x) dx = F (b) F (a) b a f(x) dx as Here is a list of basic facts about anti-derivatives that we should memorize: Example. Find 7 This is where we ended on Monday, January 4 5e x 0 x dx. Solution. To use the Fundamental Theorem of Calculus, we should first find F (x). To find F (x), we use the basic facts about anti-derivatives shown in Table 5.. In this example we break it down into more steps than many readers will need. F (x) = 5e x 0 x dx = 5 e x dx 0 x / dx split integral, and use x = x / = 5e x 0 x3/ 3/ basic anti-derivatives, adding to power = 5e x 0 3 x3/ simplifying Now we evaluate the integral at 7 and F (7) F () = 5e 7 0 ( 3 73/ 5e 0 ) 3 3/ 5.5 U-substitution = 5e / 5e Here is a brief outline of the technique of U-substitution. 0. You are given an integral h(x) dx where h(x) is some complicated function of x.. Fill in the following u = something du = (derivative of u) dx (you get to pick this) (you don t get to pick this) Most often you pick u to equal something inside of another function.. Fill in the following translate or cancel x s and dx h(x) dx = = f(u) du. Make sure that all the x s (including dx) cancel by the last step; whatever you re left with, call it f(u).

6 CHAPTER 5. INTEGRALS REVIEWED 6 3. Find the anti-derivative f(u) du = F (u) 4. Inside of F (u), replace u with the same something involving x, that you picked in step. Comments. When you are practicing u-substitution here are some good bits of advice and rules of thumb to keep in mind. To set u = something you should be willing to take a guess, try setting u equal to something, and then take the derivative, and see if you can get rid of all the x s using u and du. Above I wrote translate or cancel all the x s and dx. Different people have different approaches to this step, but they all produce the same result; approach it whichever way you want. Here s a brief description of two approaches:. just circle those x s that are part of u and those that are part of du, and replace those parts with u and du,. solve for dx, and replace it with a formula involving du and x s, and then cancel any remaining x s. Whichever approach you take, you should get h(x) dx = f(u) du and should be able to double check your work as follows. If you start with the right hand side, and substitute x s back in for u and du, you should get the original integral, the one on the left. When using u-substitution, it helps to know what sort of targets you might have for F (u) du. In other words, you want to have a list of integrals that are known, and then see if you can turn h(x) dx into one of known integrals. Example. Find our last basic functions.) tan(x) dx. (Note: this gives us the anti-derivative of one of Solution. We start by rewriting the function as a fraction: sin(x) cos(x) dx. The obvious choices for u (and the resulting du) are u = sin(x) du = cos(x) dx u = cos(x) du = sin(x) dx ( sin(x) The original integral does not have cos(x) dx, it has dx since cos(x) cos(x) dx = ) sin(x) cos(x) dx. Thus, we need to use u = cos(x) and du = sin(x) du. Now

7 CHAPTER 5. INTEGRALS REVIEWED 7 we should look at the original integral and see if we can find how to translate all the x s. Here s how it looks ( ) du sin(x) dx cos(x) }{{}. u If you replace the indicated parts shown with u and du, you should get this sin(x) cos(x) dx = u du. You should be able to double check that we ve done our work right: if you substitute back cos(x) in place of u, and sin(x) dx in place of du, then you get the original integral. Now we finish up: u du = ln u + C = ln cos(x) + C Add this to your list of basic anti-derivative facts tan(x) dx = ln cos(x) + C = ln sec(x) + C cos(ln(x)) Example. Find dx x Solution. Your intuition should be that u can equal what is inside of cosine. u = ln(x) du = x dx u {}}{ du cos( ln(x) ) dx x = cos(u) du Example 3. Find 7 sin(x) cos(x) + cos (x) = sin(u) + C = sin(ln(x)) + C Solution. Some guesses you could make in solving this integral dx u = sin(x) u = cos(x) u = cos (x) u = + cos (x) u = sin(x) cos(x) u = sin(x) cos(x) You do not need to instantly see which of these choices is best, but you would figure it out by looking at du in each case (i.e. the derivative of what u equals). Then, you would ask yourself, if you replace parts of the original integral with u and du,

8 CHAPTER 5. INTEGRALS REVIEWED 8 would you get an integral that you can finish? In other words, would an integral that is on our short list of basic anti-derivatives (c.f. page 9)? Hopefully you see that u = + cos (x) works. Here s how u = + cos (x) du = sin(x) cos(x) dx f(u) du be du = sin(x) cos(x) dx 7 sin(x) cos(x) + cos (x) du = dx 7 u du = 7 ln u = 7 ln( + cos (x)) + C The next example shows a technique called backwards substitution. It starts the same as before, identify g(x) as some part of your function, set u = g(x), take du = g (x) dx. But the last step, where we replace all x s with u s, is harder: there is an x left over after we do the usual steps. To get rid of this x, we substitute backwards: instead of having u equal to a bunch of x stuff, we ll solve this and find x equal to a bunch of u stuff. Example 4. Find x(3x + 0) 99 dx Solution. u = 3x + 0 du = 3 dx dx = 3 du x(3x + 0) 99 dx = xu 99 3 du The problem is, there appears to be no way to cancel that other x. To get rid of it we need backwards substitution. We have an equation u = 3x + 0 and we can solve this equation for x! u = 3x + 0 x = (u 0) 3 xu 99 3 du = 3 (u 0)u99 3 du = (u 0)u 99 du 9 = u 00 0u 99 du 9

9 CHAPTER 5. INTEGRALS REVIEWED 9 = ( u 0 0u00 ) = ( ) (3x + 0) 0 (3x + 0) This is where we ended on Tuesday, January 5

10 CHAPTER 5. INTEGRALS REVIEWED 0 Table 5.: Basic anti-derivatives Cf(x) dx = C f(x) dx x n dx = xn+ + C if n n + e x dx = e x + C cos(x) dx = sin(x) + C sin(x) dx = cos(x) + C f(x) ± g(x) dx = f(x) dx ± sec (x) dx = tan(x) + C dx = ln(x) + C x + x dx = tan (x) + C g(x) dx

11 Chapter 6 Applications of the definite integral All the applications in this chapter can be understood through the following principle: Approximation Principle Anything that can be approximated by adding together lots of values of a function can be found exactly by integrating. More specifically, if we re adding things like f(x) x, then we integrate f(x). Examples of what the thing being approximated could be: area between a curve and the x-axis, or area between two curves, or a volume generated by revolutions, or arc-length or surface area, or probability, etc. 6. Area between two curves Example. We want to find the area between h(x) = x + 3 x and g(x) = 3 x + (as shown to the side). We can approximate it using four rectangles (as 3 shown to the side). (a) Figure out how to find the area of one of the rectangles. (b) Use this to see the area of all four can be written as a sum of things of the form f(x i ) x. What is f(x)? (c) Use the Approximation Principle to set up an integral for the exact area, and then solve this integral. Solution. It appears that each rectangle has width x =. We ll pick the third rectangle to look at. It looks like it s centered at x =.5. The top edge has height h(.5) and bottom edge is at g(.5). Thus, it s height is h(.5) g(.5), and Area of third rectangle = (h(.5) g(.5)) The area of all four rectangles can be written as (h(.5) g(.5)) +(h(.75) g(.75)) +(h(.5) g(.5)) +(h(.75) g(.75))

12 CHAPTER 6. INTEGRAL APPLICATIONS Thus, we have we have a sum of things of the form (h(x) g(x)) x, i.e. f(x) = h(x) g(x). Since we have figured out what to sorts of things to add to approximate the area, we can now apply the Approximation Principle and integrate this instead: 3 A = h(x) g(x) dx = 3 x3 + x 3 3x = 4 3 Definition. The area above one curve and below the other is given by Definition. If f(x) g(x), then In general, Area = b a top curve bottom curve dx area between f(x) and g(x) and between x = a and x = b area between f(x) and g(x) and between x = a and x = b = = b a b a f(x) g(x) dx f(x) g(x) dx To calculate an integral of the second kind, we need to split the integral up into two (or more) pieces such as b f(x) g(x) or only g(x) f(x). a = c a + b c where in each piece we have only This is where we ended on Wednesady, January 6 Comments. There are a few variations on this type of problem. Sometimes you have to solve for a and b if they have not been given explicitly. Sometimes it changes which function is on top and which is on the bottom. In this case, you need to split the integral into two pieces, corresponding to the change. Finally, sometimes the area is defined left-to-right instead of to top-to-bottom. The Approximation Principle can also work for areas that are to the left of one curve and to the right of the another curve. In this case, you can picture sideways rectangles, and think of everything as a function of y. Thus, you ll have... dy. Example. The shapes below are defined by the functions x = y + 5 and x = y y + 0. (a) Find the measurements of the rectangle shown below. (b) Find the y-values for the intersections of the two curves.

13 CHAPTER 6. INTEGRAL APPLICATIONS 3 (c) Find the area shown between the straight line and the parabola. Solution. (a) The rectangle appears to be lined up with y-values of and 3. Thus, the height is. The width is the difference between the x-values on the line and the parabola. The x-values correspond to y = 3, thus we have x (3) x (3) = (3) + 5 ( ) = 6 (b) To find the points of intersection, we solve and get y + 5 = y y + 0 y = 5 ± , (c) To find the area, we imagine adding together the area of rectangles of the sort we had in part. This means we take the integral of such rectangles. The value that changes is the width of the rectangle, and this is given by subtracting x (y) x (y), i.e. by integrating y + 5 (y y + 0). Now we can use the values from part (b) and finish the problem: x (y) x (y) dy = = y + 5 (y y + 0) dy y + 0y + 5 = y y + 5y = ( 0.477) + 5( 0.477) ( ) + 5(0.477) + 5(0.477) Definition. The area to the right of one curve and to the left of another curve equals Definition. If f(y) g(y), then Area = y=b y=a right curve left curve dy area between f(y) and g(y) and between y = a and y = b = y=b y=a f(y) g(y) dy.

14 CHAPTER 6. INTEGRAL APPLICATIONS 4 Extra examples Example 3. Find the area between f(x) = x and g(x) = x + x + 5 (shown to the side), and illustrate four rectangles in the Riemann Sum corresponding to the integral you use. Solution. To find a and b we intersect f(x) = x and g(x) = x + x + 5: x x 5 = 0 x = x + x + 5 x = ± 4 4()( 5) 4 x = ± x.58,.58 Note that g(x) is on top, i.e. g(x) f(x). Thus, our integral is x + x + 5 x dx. We simplify and calculate this integral.58 x + x 5 dx = 3 x3 + x 5x A four step Riemann Sum is pictured below b a g(x) f(x) dx (g(x ) f(x )) x + (g(x ) f(x )) x + (g(x 3 ) f(x 3 )) x + (g(x 4 ) f(x 4 )) x Area Comments. Note: the type of justification given at the end of the previous example may seem shallow and/or unnecessary, but the point is this: the student should not need to memorize each integral formula that we learn. Rather, we find each of these formulas in exactly the same way. Start by asking, can we break a quantity down into pieces of the form f(x i ) x? If so, then the whole quantity is found by integrating, f(x) dx. The whole quantity in the previous example can be breaking down the area into pieces of the form ( g(x i ) f(x i ) ) x.

15 CHAPTER 6. INTEGRAL APPLICATIONS 5 Example 4. Find the area that is between f(x) = sin(x) and g(x) = 0.5x and between x = 0 and x = 5, as shown below (use your calculator to find the intersection in the middle, then split the integral into two parts). Solution. The endpoints are 0 and 5. We do not always have one function on top, so we need to use absolute values; the area is 5 0 sin(x) 0.5x dx. To solve this we need to split the integral up: one part will have sin(x) on top and one part will have 0.5x on top. The splitting up occurs at the point of intersection. To find the point of intersection we can use our calculator, either using cnd, then CALC, then 5: intersect, or just using zoom and trace. The intersection is at x.475. Thus our integral becomes sin(x) 0.5x dx + = sin(x) 0.5x dx sin(x) 0.5x dx x sin(x) dx = ( cos(x) 0.5x ) ( 0.5x + cos(x) ) (Note: you can double check your answer using your calculator: enter fnint(abs(sin(x) - 0.5x),x,0,5).) 6. Volumes by rotation The most basic shape for volumes is rectangular: On the Macintosh computer, you can use the Grapher application find the intersection. Go to the /Applications folder, then the /Utilities folder. Open Grapher, enter one of the functions in the first y = field, then go to the menu for Equation New Equation to enter the second one. Select both the equations, then go to Equation Find intersection..475

16 CHAPTER 6. INTEGRAL APPLICATIONS 6 y z x Volume = x y z We can generalize this: any shape that has constant cross section (i.e. the same shape in for all slices ) with area A, and length l, has an easy volume formula: Cross section: A l Volume = A l volume of a shape with constant cross-section of A is A l where l is the length of the shape. The most important case, for us, of this volume formula is for a cylinder: r h Volume = πr h And, the most important application of the cylinder is where we have thickness x x r Volume = πr x On Friday, we got through solving part (a) of the next example Example. Imagine we want to find the volume of hard boiled egg. We could put the egg in a measuring cup and measure how much water it displaces. But we suppose we want to do it more mathematically, using functions and formulas. We will imagine cutting the egg into slices, and then measuring the volume of each slice as a cylinder. (a) Use the ellipse x.55 + y 0.8 = to model the outline of an egg with length.3 in and diameter.6 in. Draw the ellipse and the shape it makes when rotated around the x-axis. (b) Imagine cutting the shape from (a) into slices, and draw the result. (c) Imagine replacing each slice from (b) with a cylinder and draw the result.

17 CHAPTER 6. INTEGRAL APPLICATIONS 7 (d) Figure out how to find the volume of each cylinder-slice in (c) and write down a formula for the sum of the volumes. (e) Translate the formula from (d) into an integral, and solve the integral. Solution. (a) We imagine the egg being the three dimensional solid defined by rotating this ellipse around the x-axis: (b) Show how we can slice this volume into pieces, and approximate the volume of each piece using a disk, and then turn the the sum of volumes of disks into an integral. We start by drawing a picture of the egg after it s cut into a bunch of slices (notice that cutting a shape into pieces doesn t change the total volume): Vol (c) Now each slice has a slight curve on the top and bottom. This makes it hard to find the exact volume of the slice, but we can approximate the volume by using a cylinder of the same size: Vol (d) Now, we can give a formula for the volume of all these disk-shaped slices above stuff = πr x + πr x + πr 3 x +...

18 CHAPTER 6. INTEGRAL APPLICATIONS 8 To find r, r, etc., we go back to how this shape is defined: the radius is the distance of the curve from the x-axis. This distance is what we usually call the x y-value, i.e. r = y. Now we find the y-values from the ellipse.55 + y 0.8 = by solving for y: y 0.8 = x.55 ( ) y = 0.8 x.55 ( ) y = 0.8 x.55 y = 0.8 x.55 Combining this formula with the above sum of volumes we get ( ) ( ) ( ) above stuff = π 0.8 x /.55 x+π 0.8 x /.55 x+π 0.8 x 3 /.55 x+... (e) Now we apply the Approximation Principle: anything that we can approximate with a sum of function values, we can make exact by integrating: V = ( π 0.8 ) x /.55 dx This last line is the most important conclusion of this example: translating a volume by rotation into an integral. Of course, it s also nice to find the integral: π (0.8 ).55 x /.55 dx = π 0.8 ( x /.55 ) dx = π(0.8) x /.55 dx.55 = π(0.8) (x = 3. in 3 x 3 ).55 3(.55).55 This is where we ended on Friday, January 8 Rule. Let V be the volume generated by rotating f(x) around the x-axis, between x = a and x = b. Then V is given by the following formula: V = b a πr dx, where r = f(x). Example. Find the volume generated by rotating y = x + around the x-axis, between x = and x =. Solution. The original shape, and the volume generated by it are pictured below.

19 CHAPTER 6. INTEGRAL APPLICATIONS 9 r = y = x + x We apply V = b a π(f(x)) dx with a =, b =, f(x) = x + and get π( x + ) dx = π x 4 4x + 4 dx ( = π 5 x5 4 ) 3 x3 + 4x ( = π 5 5/ 4 3 3/ + 4 ( 5 5/ / 4 )) ( = π 5 5/ 4 3 3/ + 4 ) ( = π ) 5 3 = 4 π ( 5 ) 3 + = 8 ( 3 π ) 5 = 64 π 5 Example 3. Find the volume generated by rotating y = x around the y-axis, between y = 0 and y =. Solution. This is a sideways problem because the way we can cut it into disks is to make cuts that go up and down along the y-axis.

20 CHAPTER 6. INTEGRAL APPLICATIONS 0 One slice of this volume gives a disk of thickness y and and radius given by the x-value. To get the x-value we need to write x(y), i.e. x as a function of y. We solve y = x for x = y. We also need to know the bounds of integration; this is from the smallest y-value to the largest one. The y-values are y = 0 and y =. 0 π( y) dy = π 0 = π y = π = π y dy 0 Rule (Generalized from previous example). The volume generated by rotating f(y) around the y-axis, between y = a and y = b is Volume of rotation = b a πr dy, where r = f(y) This is where we ended on Tuesday, January Example 4. Set up an integral, and use your calculator to find it, for the volume of the napkin ring made by rotating the region bounded by y = x +, and y = around the x-axis. Solution. We picture the region defined below, along with one slice of volume: The volume of the slice is given by the area of the face, times x. The main work is to figure out the formula for the area. We start with the two dimensional shape, involving r and R

21 CHAPTER 6. INTEGRAL APPLICATIONS r R Area = πr πr Volume = (πr πr ) x So, to finish, we just need to figure out formulas for R and r in the shape that we have. R r Thus, R = y = x + and r =. Putting all of this together, we have the integral that we want. V = π ( x + ) π dx V = = = π = π π ( x + ) π dx π( x + 4 x + 4 4) dx ( x + 4 x + dx To finish this, we need to figure out what ) x dx x dx is. This is probably not something that you know how to find the anti-derivative of. But, you can figure out the integral without knowing the anti-derivative. If you think of what area is

22 CHAPTER 6. INTEGRAL APPLICATIONS represented by x dx, you should be able to see that it s the area of the top half of a unit circle. This has area π. Thus, ( 4 V = π ) π = 4 3 π + π Comments. We have seen at least three variations on rotation so far: one function around the x-axis, one function around the y-axis, and two functions around the x-axis. There are a few more variations, but rather than introducing each one separately, it might be nice to see all the variations all at once: Rule (Volumes of rotation by disk and washer). Disks and washers around horizontal lines disks washers V = πr dx V = πr πr dx around x-axis r = f(x) R = f(x), r = g(x) around y = c r = f(x) c or c f(x) R = f(x) c or c f(x), r = g(x) c or c g(x) (choose the formulas for R and r that are positive, and with R > r) Disks and washers around vertical lines disks washers V = πr dy V = πr πr dy around y-axis r = f(y) R = f(y), r = g(y) around x = c r = f(y) c or c f(y) R = f(y) c or c f(y), r = g(y) c or c g(y) (choose the formulas for R and r that are positive, and with R > r) Example 5. Find the volume of the napkin ring made by rotating the region bounded by y = 4 9 x +, and y = 4 x + 7 (shown below) around the line 6 y =.

23 CHAPTER 6. INTEGRAL APPLICATIONS 3 Solution. The rotated shape looks roughly as follows: The larger radius is defined by the top curve, y = 4 9 x +. It is the distance between this curve and the horizontal axis, y =. Thus, R = 4 9 x + ( ) = 4 9 x + 3. The smaller radius, the radius of the hole, is defined by the bottom curve y = 4 x + 7. It is the distance between this curve and the horizontal axis y =. 6 Thus r = 4 x ( ) = 4 x Now we integrate (skipping some of the messy steps) V = 3/ 3/ ( π 4 ( 9 x + 3) π 4 x + 3 ) dx 6 (messy steps of foiling skipped) = π = 5π 6 = 5π 6 3/ 3/ 3/ x x dx 3/ 7 8 x4 3 6 x dx ( x5 3 8 x x ) 3/ 3/ = 5π = π This is where we ended on Wednesday, January 3

24 CHAPTER 6. INTEGRAL APPLICATIONS 4 Example 6. Find the volume of the region between the curves y = g(x) = x 3 and y = f(x) = x rotated around the line x = (shown below). Solution. We solve this with washers, that are stacked up and down along the vertical line x =. Thus, we will have b a πr πr dy where we need R and r to be formulas for the radiuses, written as functions of y. The larger radius, R, is the horizontal distance between x = and the curve f(x). We need the formula in terms of y, so we solve y = x for x = y, R = y The smaller radius, r, is the horizontal distance between x = and the curve g(x). Again we rewrite the equation first y = x 3 x = y /3 Putting it all together we get r = y /3

25 CHAPTER 6. INTEGRAL APPLICATIONS 5 Extra examples V = = π 0 0 = π 0 ( = π π( y ) π( y /3 ) dy y + y 4 ( y /3 + y /3 ) dy y + y 4 + y /3 y /3 dy ) 3 y3 + y5 y4/ /3 y5/3 5/3 ) ( = π = 3 30 π Example 7. Derive the formula for the volume of a sphere of radius r. Solution. We start with the formula for the top half of a circle of radius r: 0 y = r x and then plug this into our volume by rotation formula V = = r r r r π( r x ) dx π(r x ) dx ) = π (r x x3 r 3 = π (r r r3 = π (r 3 r3 3 = π 3 r3 = 4 3 πr3 3 ) r ( r r + r3 3 )) Challenge. Can you figure out how to find the volume of a shape rotated around the line y = x? What about other lines? Can you figure out how to apply washers when f(x) and g(x) switch places with respect to which one is farther from c? What about if they switch places also with respect to which side of c they fall on?

26 CHAPTER 6. INTEGRAL APPLICATIONS 6 Pictures of areas and volumes f(x) g(x) f(x) g(x) = h x Area of this rectangle = (f(x) g(x)) x Area between = two curves f(x) g(x) dx f(x) g(x) } {{ } } y g(x) f(x) Area of this rectangle = (g(y) f(y)) y Area between two sideways curves = g(y) f(y) dy

27 CHAPTER 6. INTEGRAL APPLICATIONS 7 f(x) r = f(x) x Volume of this disk = π(f(x)) x Volume of revolution between one function and the x-axis, using disks = π(f(x)) dx

28 CHAPTER 6. INTEGRAL APPLICATIONS 8 f(x) g(x) R r x-axis x Volume of this washer = (πr πr ) x = ( π(f(x)) π(g(x)) ) x Volume of rotation between two functions, rotated around the x-axis, using washers π(f(x)) π(g(x)) dx =

29 CHAPTER 6. INTEGRAL APPLICATIONS 9 f(x) g(x) line y = c x-axis R r line y = c x Volume of this washer = (πr πr ) x = ( π(f(x) c) π(g(x) c) ) x Volume of rotation between two functions, rotated around the line y = c, using washers π(f(x) c) π(g(x) c) dx = 6.3 Volumes by Cylindrical Shells For some functions it s easier to slice the volume a different way than in the previous section. In particular, we ll add one more basic volume shape to our repertoire:

30 CHAPTER 6. INTEGRAL APPLICATIONS 30 cylindrical shells. One is pictured below r h r To figure out this volume you can imagine cutting the shell and unrolling it and flattening it out, as shown below in four steps. The result is a rectangular solid with measurements r, h and πr, and so the volume is V (cylindrical shell) = πrh r h πr r This is where we ended on Friday, January 5 Example. Find the volume obtained by rotating the region between y = sin(x ) and y = 0, from x = 0 to x = π around the y-axis. Solution. We show the original area and the volume of rotation it generates below. We can picture this volume as being cut into cylindrical shells. We show below how this looks, both showing first all the shells at the same time, and then one shell at a time.

31 CHAPTER 6. INTEGRAL APPLICATIONS 3 Note, there is no way to do this problem (easily) with disks or washers. If we used them we d need to do it sideways using functions of y. But solving y = sin(x ) for x gives x = sin (x) and we don t know how to integrate this. Thus, the total volume can be found by adding volumes of the form πr xh. The integral form is V = πrh dr We replace r, x and h with r = x, r = x, and h = f(x). Thus, our integral becomes Let u = x, du = x dx. V = = π π 0 π 0 πx sin(x ) dx x sin(x ) dx x=π V = π sin(u) du x=0 x=π = π cos(u) x=0 = π cos(x x=π ) x=0 = π ( cos(π) cos(0) ) = π( ) = π This is where we ended on Monday, January 8

32 CHAPTER 6. INTEGRAL APPLICATIONS 3 Rule (Volumes of rotation by cylindrical shells). Cylindrical shells around vertical lines One function Two functions b a πrh dx b a πrh dx around y-axis r = x, h = f(x) r = x, h = f(x) g(x) around x = c r = x c or c x, h = f(x) r = x c or c x, h = f(x) g(x) (choose the formula for r that is positive; make sure f(x) is on top of g(x)) Example. Sketch the region and one typical cylindrical shell for the volume generated by the region between y = sin(πx) and y = x x +, and from 4 a and b 0.938, rotated around the line x =.5. Then set up (but do not solve) an integral for this volume. Solution. Here is a picture of the original region, and a shell in the rotated volume.

33 CHAPTER 6. INTEGRAL APPLICATIONS 33 As usual, we wish to integrate volumes of the form πrh x. As stated in our rule, h is found by subtracting one curve from the other h = f(x) g(x) with f(x) = sin(πx) and g(x) = x x +. Also, we find r by subtracting. It is the 4 distance between a point on the curve and the line x =.5. This distance is found as.5 x. Thus, we have the following integral V = π(.5 x) (sin(πx) (x x + 4 ) ) dx Average Value of a Function Example. Suppose that on one day the temperature is given by the following function K(t) = 0.4t 45 sin(πt/) where t is in hours (with t = 0 corresponding to midnight), and K(t) is the temperature in Fahrenheit. (a) Find a formula for the average value of the temperature for the whole day by using 4 equally spaced times. (b) Find a formula for the average value of the temperature for the whole day by using equally spaced times. (c) Set up and solve an integral for the average value of the whole day. Solution. (a) We will use t = 0, 6,, 8. Then the average is A = K(0) + K(6) + K() + K(8) 4 Note that the formula is what we are most interested in here, not the actual calculation. However, it might be nice to see the results F (b) We will use t = 0,, 4, 6, 8, 0,, 4, 6, 8, 0,. Then the average is A = K(0) + K() + K(4) + + K(6) + K(8) + K(0) + K() (6.) As above, we are most interested in the formula, but also include the actual calculation: F (c) The key to turning the above calculations into an integral are to see them as sums of things of the form K(t i ) t. It s pretty clear how to see K(t i ) in the above sums, but we need to rewrite things a little to see where t shows up. In part (b) we divided by, the number of steps we were asked to use. Using n = steps is equivalent to having t = ; in terms of an equation this means = 4

34 CHAPTER 6. INTEGRAL APPLICATIONS 34 n = t 4 Thus, our calculation in Equation (6.) is equivalent to t 4 (K(0) + K() + + K()) = (K(0) t + K() t + + K() t) 4 If we use larger values of n, and eventually let n, the 4 doesn t change, but the sum on the right changes into an integral. Thus, the exact average is given by the following integral: Now we find this integral A = 4 = 4 = 4 A = K(t) dt. 0.4t 45 sin(πt/) dt ( 0.t + 45 cos(πt/) ) 4 π 0 ( 0.t + 45 cos(πt/) ) 4 π = 4 5. = This is where we ended on Tuesday, January 9 Definition (Generalized from previous example). The average value of a function f on an interval [a, b] is f avg = b f(x) dx b a Comments. If f(x) 0 then we can understand this definition another way: geometrically. If we think of b width, then we can rewrite this definition: a a f(x) dx in terms of area, and b a in terms of area = average height width b a f(x) dx and so f avg = b a average height = area width becomes Example. Find the average value of the function f(x) = x 3x on the interval [, ]. Solution. f avg = x 3x dx = ( ) x x3 =

35 Chapter 7 Techniques of Integration 7. Integration by Parts Integration by parts is doing the product rule backwards, i.e. taking the antiderivative of the product rule. (f g) = f g + f g product rule (f g) = (f g + f g ) taking of both sides f g = f g + f g simplifying Now we solve the last equation for one of the remaining integrals to get the formula we want: Rule. Integration by parts f g = f g f g This is where we ended on Wednesday, January 30 Comments. Comments. When I forget this formula, I usually write down one or two lines from the above calculations starting with the product rule, and then get the result again. Comments. The book writes this a different way. Let v = f(x) and u = g(x) so dv = f (x) dx and du = g (x) dx. Then we have the following rule. Rule. Integration by parts (alternative notation) This formula has a mnemonic: You devil, ultra-violet voodoo: u dv = u v u y o u dv d e v i l v du = u v u v l i t o r l a e t v v o o du d o o Note,when you use this formula, you have been given an integral like xe x dx. There are two functions in this integral, however neither of them looks like f, that is to say, neither looks like a derivative. It s up to us to pretend that one of them is the derivative of something. 35

36 CHAPTER 7. TECHNIQUES OF INTEGRATION 36 Comments. I urge students to try a little bit to look past the notation of the integration by parts. In other words, it s OK to use f s and g s or u s and v s, and it s OK to really prefer one over the other. But, also look past what all these letters are and think about what the formula is saying in words. Rule. Integration by parts, verbally two functions = original anti-deriv deriv anti-deriv Comments. When you use the verbal description of Integration by Parts, just remember that you use the same function for both anti-derivatives, and use the other function for the original and derivative spots. Comments. Usually you are given something to integrate that looks like a product. You have to choose which thing to call f (or du) and which to call g (or v). The point is that the second integral should usually be easier for some reason than the original integral. Example. Find xe 3x dx. Solution. This time we will use the formula u dv = uv v du. Our original integral has x and e 3x. We need to choose which of these functions to call u and which to call dv. We make this choice so that the integral v du is simpler; in particular, we should choose u so that du will be simpler. To do this, we choose u = x and therefore dv = e 3x dx. Thus, we fill in the following u = x du = dv = e 3x dx v = 3 e3x (choose this) (fill in this) and we put the results together using the IBP formula, xe 3x dx = uv v du = 3 xe3x 3 e3x dx, and finish by taking the last anti-derivative, Heuristic 3 xe3x 3 e3x dx = x 3 e3x 9 e3x. Comments. One question students often have is which function in the integral to call f and which to call g, or, equivalently, which to call u and which to call dv, or, equivalently, which to take the derivative of, and which to take the anti-derivative of.

37 CHAPTER 7. TECHNIQUES OF INTEGRATION 37 Rule. One rule of thumb is to take the derivative of the function (i.e. set g or u equal to the function) that shows up first on the following list: LIATE: Logarithmic, Inverse trig, Algebraic, Trig, Exponential. Example. Find t sin(5t) dt. Solution. Using the formula as shown: g = t f = sin(5t) (choose this) f g = fg g = t fg we make our choices and fill in f = 5 cos(5t) (fill in this) Now we plug the results into our IBP formula: ( t sin(5t) dt = t ) 5 cos(5t) t ( 5 ) cos(5t) dt. Now we rewrite the last integral a little as + 5 We let g equal the algebraic function: t cos(5t) dt and apply IBP again. g = t g = f = cos(5t) (choose this) f = 5 sin(5t) (fill in this) and plug the IBP formula into the integrals above above stuff = 5 t cos(5t) + ( 5 5 t sin(5t) Example 3. Find = 5 t cos(5t) + 5 for the equation for the missing integral.) 5 ( 5 t sin(5t) + 5 cos(5t) ) sin(5t) dt ) e x sin(x) dx. (Hint: do integration by parts twice, and solve Solution. Let f = e x and g = sin(x). Then f = e x and g = cos(x). e x sin(x) dx = e x sin(x) e x cos(x) dx Now we apply integration by parts again. Let f = e x and g = cos(x). Then f = e x and g = sin(x). e x sin(x) dx = e x sin(x) e x cos(x) dx ( ) = e x sin(x) e x cos(x) e x ( sin(x)) dx

38 CHAPTER 7. TECHNIQUES OF INTEGRATION 38 The equation is now e x sin(x) dx = e x sin(x) e x cos(x) e x sin(x) dx We solve this for the unknown integral e x sin(x) dx = e x sin(x) e x cos(x) e x sin(x) dx = ex (sin(x) cos(x)) Example 4. Find This is where we ended on Friday, February ln(x) dx (the last of our basic anti-derivative). Solution. The trick here is that to use integration by parts, we really, really need to view ln(x) as a product. That way we can choose to view it as f g. Can you think of how to make ln(x) into a product? We write the original integral as ln(x) dx. Then we choose f = and g = ln(x). From this we get f = x and g = x. ln(x) dx = ln(x) dx = x ln(x) = x ln(x) x dx x dx = x ln(x) x ln(x) dx = x ln(x) x Example 5. Find tan (x) dx. Solution. This problem is like the previous one; we apply integration by parts using one factor equal to. tan (x) dx = tan (x) dx Now we pick f and g, and calculate f and g f = f = x g = tan (x) g = + x Now we apply these formulas using integration by parts

39 CHAPTER 7. TECHNIQUES OF INTEGRATION 39 tan (x) dx = tan (x) dx = x tan (x) x + x dx = x tan x (x) + x dx To figure out this last integral you can use u-substitution. Now we can finish the integral x + x dx = u du = ln + x u = + x du = x dx tan (x) dx = x tan (x) ln + x Tabular integration by parts Comments. An integral of the form x f(x) dx or x 3 f(x), etc., can involve repeated integration by parts. Whenever an integration by parts is repeated, you might get the sense that a pattern is emerging as you do the steps. However, it can be hard to see the pattern after you ve written all those integral symbols, and f s and g s, etc. Tabular integration by parts reveals this pattern. We simplify the notation a little, by writing this formula using f and g (as opposed to f and g): Rule (Tabular Integration by Parts). fg = g f g f + g f g f +... where the terms on the right equal the functions connected by arrows below +g g f (both original functions) g f f f g Example 6. Find x 3 sin(πx) dx

40 CHAPTER 7. TECHNIQUES OF INTEGRATION 40 Solution. We make two columns of functions g f +x 3 sin(πx) 3x π cos(πx) +6x π sin(πx) 6 π 3 cos(πx) 0 π 4 sin(πx) Now we take the terms that are connected by arrows in the above tables. x 3 sin(πx) dx = x3 3x 6x cos(πx) + π π sin(πx) + π 3 cos(πx) 6 π 4 sin(πx) x 3 3x + 0 Example 7. Find dx. x 5 Solution. We start by viewing the integral as the product of two functions x 3 3x + 0 dx = (x 3 3x + 0) (x 5) / dx. x 5 We make two columns of functions g f x 3 3x + 0 (x 5) / (both original functions) (6x 3) (x 5) / / = (x 5) / (x) (x 5) 3/ 3/ = 3 (x 5)3/ () (x 5) 5/ 3 5/ = 3 5 (x 5)5/ 0 3 (x 5) 7/ 5 7/ = (x 5)5/ Now we take the terms that are connected by arrows in the above tables. x 3 3x + 0 x 5 dx = (x 3 3x + 0)(x 5) / 3 (6x 3)(x 5) 3/ + (x) 3 5 (x 5)5/ () (x 5)7/ This is where we ended on Monday, February 4 7. Trigonometric Integrals Comments. This section gives tricks for solving integrals of the form and tan n (x) sec m (x) dx. sin n (x) cos m (x) dx

41 CHAPTER 7. TECHNIQUES OF INTEGRATION 4 Example. Find sin(x) cos 8 (x) dx. Solution. Let u = cos(x), du = sin(x) dx and the integral becomes u 8 du Then the anti-derivative is u9 9 = cos9 9 Comments. The same basic approach that we used in the previous example will work with any power of cosine. We could have had cos (x), cos 5 (x), or even cos(x). In fact, we can even extend the same idea to any integral with an odd number of sine functions, provided we use a trig identity to get rid of the other powers. A similar trick will work if we have an odd number of cosine functions. We give an outline of this procedure first, and then do another example. Rule. To integrate sin n (x) cos m (x) dx: Case If the power of sine is odd, then let u = cos(x), du = sin(x) dx, and get rid of all but one power of sin(x) using sin (x) = cos (x) sin 4 (x) = ( cos (x)) sin 6 (x) = ( cos (x)) 3 etc and then complete your u-substitution and integrate. Case If the power of cosine is odd, then let u = sin(x), du = cos(x) dx, and get rid of all but one power of cos(x) using cos (x) = sin (x) cos 4 (x) = ( sin (x)) cos 6 (x) = ( sin (x)) 3 etc and then complete your u-substitution and integrate. Case 3 If both sine and cosine have even powers, then use the identities sin (θ) = ( cos(θ)) cos (θ) = ( + cos(θ)) then multiply everything out. You now have only powers of cos(x). Split the integral up: odd powers bigger than, go to Case ; even powers, repeat Case 3. In this way you are eventually left with only single powers of cos(x), cos(4x), cos(8x),..., which you can finish immediately. This is where we ended on Tuesday, February 5

42 CHAPTER 7. TECHNIQUES OF INTEGRATION 4 Example. Find sin 7 (x) cos(x) dx. Solution. Let u = cos(x), du = sin(x) dx. We get rid of sin 6 (x) by rewriting it as ( cos (x)) 3. Then we have: sin 7 (x) cos(x) dx = sin(x)( cos (x)) 3 (cos(x)) / dx = ( u ) 3 u / du = ( 3u + 3u 4 u 6 )u / du = u / 3u 5/ + 3u 9/ u 3/ du ( = 3 u3/ 6 7 u7/ + 6 u/ ) 5 u5/ ( = 3 (cos(x))3/ 6 7 (cos(x))7/ + 6 (cos(x))/ ) 5 (cos(x))5/ Example 3. Find sin (x) cos (x) dx Solution. We apply the identities mentioned in Case 3 above to get ( cos(x)) ( + cos(x)) dx = cos (x) dx 4 = ( + cos(4x)) dx 4 = 4 cos(4x) dx = cos(4x) dx 8 = ( x sin(4x) ) 8 4 This is where we ended on Wednesday, February 6 Comments. Now we apply the exact same ideas to integrals of the form This time, we start with an outline of the approach. Rule. To integrate tan n (x) sec m (x) dx Case If the power of tangent is odd get then let u = sec(x), du = sec(x) tan(x) dx, and get rid of all but one power of tan(x) using tan (x) = sec (x) tan 4 (x) = (sec (x) ) tan 6 (x) = (sec (x) ) 3 etc and then complete your u-substitution and integrate. tan n (x) sec m (x) dx.

43 CHAPTER 7. TECHNIQUES OF INTEGRATION 43 Case If the power of secant is even then let u = tan(x), du = sec (x) dx, and get rid of all but two powers of sec(x) using sec (x) = tan (x) + sec 4 (x) = (tan (x) + ) sec 6 (x) = (tan (x) + ) 3 etc and then complete your u-substitution and integrate. Case 3 If tangent has an even power and secant an odd power, then get rid of all the powers of tan(x) using tan (x) = sec (x) as above. Now we have only powers of sec(x). Use a little luck, integration by parts, the secant-tangent identity, and the following: sec(x) dx = ln sec(x) + tan(x). Example 4. Find tan 3 (x) sec 3 (x) dx. Solution. Note that the power of tangent is odd, so we use u = sec(x) du = sec(x) tan(x) dx If we apply this we can translate part of the integral already tan (x) tan(x) sec(x) sec (x)()dx = tan (x)u du }{{}}{{}}{{} du u du Now to finish we need to get rid of the last functions of u. We do this with the identity tan (x)u du = (sec (x) )u du = (u )u du = u 4 u du Example 5. Find sec 3 (x) dx. = u5 5 u3 3 = sec5 (x) 5 sec3 (x) 3 Solution. We have an even number of tangents (zero) and an odd number of secants. The integral is already written in terms of secants, so we apply integration by parts. To do this, we need to write sec 3 (x) as a product, namely as sec(x) sec (x). anti-derivative {}}{ sec(x) sec (x) dx = sec(x) tan(x) derivative anti-derivative {}}{{}}{ sec(x) tan(x) tan(x) dx

44 CHAPTER 7. TECHNIQUES OF INTEGRATION 44 = sec(x) tan(x) = sec(x) tan(x) = sec(x) tan(x) sec 3 (x) dx = sec(x) tan(x) sec(x) tan (x) dx sec(x)(sec (x) ) dx sec 3 (x) + sec(x) dx sec 3 (x) dx + ln sec(x) + tan(x) sec 3 (x) dx = sec(x) tan(x) + ln sec(x) + tan(x) sec 3 (x) dx = ( sec(x) tan(x) + ln sec(x) + tan(x) ) Example 6. Find tan (x) sec(x) dx. Solution. Since we have an even number of tangents, and an odd number of secants, we start by getting rid of all powers of tangent tan (x) sec(x) dx = (sec (x) ) sec(x) dx = sec 3 (x) sec(x) dx. The integrals are now known: sec(x) dx was given above, and the previous example solved sec 3 (x) dx. We put this in, simplify, and we are done: ( ) sec(x) tan(x) + ln sec(x) + tan(x) ln sec(x) + tan(x) = ( ) sec(x) tan(x) ln sec(x) + tan(x) This is where we ended on Monday, February 7.3 Trigonometric Substitution The basic idea here is that we reverse the usual role of u-substitution. Usually, we set u equal to some function of x because this covers up some complicated function. But here, we re going to set x equal to a more complicated function (of θ) because of the special properties of trig functions. Example. Find the area under under y = x from x = 0 to x = /. Solution. This is the area under part of a circle, as shown,

45 CHAPTER 7. TECHNIQUES OF INTEGRATION 45 But this area is not one quarter of the half circle. There is no elementary way to find this area. The area we want is given by an integral / 0 x dx. We do a kind of backwards substitution: instead of letting u = g(x), we will let x = g(θ) where g is some trig function. We will soon give the rules of thumb for which trig function to use, but for now, as an act of faith, we do the correct substitution. Let x = sin(θ) dx = cos(θ) dθ Plugging these into the above integral we obtain x=/ x=0 sin (θ) cos(θ) dθ. We need to do two things to this integral: change the endpoints from x-values to θ-values, and simplify the formula on the inside. Here are the calculations, x = 0 sin(θ) = 0 θ = 0 x = / sin(θ) = / θ = π/6 sin (θ) = cos (θ) = cos(θ) Note where we put : this last equation is true as long as cos(θ) is positive. If cos(θ) is negative, then cos (θ) = cos(θ). The integral now becomes π/6 0 cos(θ) cos(θ) dθ = π/6 We look up this integral from section 7. and get π/6 0 cos (θ) dθ = π/6 0 0 cos (θ) dθ. ( + cos(θ)) dθ = θ + π/6 4 sin(θ) 0 = π + 4 sin(π/3) (0 + 4 sin(0)) = π The following page summarizes the main steps in Trigonometric substitution. 4 Example. Find x dx.

46 Complete Table for Trigonometric Substitution Follow the table from left to right, working in one row the whole time. Translating the integral with a substitution After the antiderivative involves Substitution original becomes sister trig function Transition Definite integral: Change endpoints from x = a and x = b Indefinite integral: Rewrite θ and other trig functions as functions of x A x x = A sin(θ) A A sin (θ) Now the original integral θ = sin (a/a) θ = sin (x/a) has become one A x involving only powers dx = A cos(θ) dθ = A sin (θ) θ = sin (b/a) of trig functions. cos(θ) = A A + x x = A tan(θ) dx = A sec (θ) dθ = A x A x First simplify this = A cos (θ) = A cos(θ) tan(θ) = as much as possible. When you re done A + A tan (θ) the integral should be θ = tan (a/a) θ = tan (x/a) known in the sense + tan (θ) that we ve done ones θ = tan (b/a) like it before and the sin(θ) = results can be looked up (even copied from = A sec (θ) = A sec(θ) cos(θ) = tables). x A + x A A + x After taking the antiderivative you finish x A x = A sec(θ) A sec (θ) A θ = sec (a/a) θ = sec (x/a) A with one or the other x θ = sec (b/a) sin(θ) = (not both) of the two x dx = A sec(θ) tan(θ) dθ = A sec (θ) columns to the right. x A = A tan (θ) = A tan(θ) tan(θ) = A

47 CHAPTER 7. TECHNIQUES OF INTEGRATION 47 Solution. This problem is very similar to the previous one: the differences are that we have 4 instead of under the square root, and we need to find a general anti-derivative, not get a definite integral and plug in numbers. x = sin(θ) dx = cos(θ) dθ Plugging these into the above integral we obtain 4 ( sin(θ)) cos(θ) dθ = 4 4 sin (θ) cos(θ) dθ = sin (θ) cos(θ) dθ cos = 4 (θ) cos(θ) dθ = 4 cos(θ) cos(θ) dθ = 4 cos (θ) dθ = 4 ( + cos(θ)) dθ = 4 (θ + sin(θ)) = (θ + sin(θ)) To finish, we just need to translate from θ back to x. This is easier if we first apply the double angle identity, sin(θ) = sin(θ) cos(θ), to get (θ + sin(θ)) = (θ + sin(θ) cos(θ)) By the original substitution, x = sin(θ). This allows us to get rid of sin(θ) directly and to get rid of θ by solving for θ = sin (x/). Plugging this in we get (sin (x/) + x ) cos(θ) (partial translation back to x) To find cos(θ) in terms of x you can draw a right triangle, label an angle as θ, the opposite side as x, the hypotenuse as (this is because sin(θ) = x/) and solve for 4 x the missing side. You should find that cos(θ) = (by the way, it always works out that the missing side is the that you started with in the integral). Thus, ( 4 x dx = (θ + sin(θ)) = sin (x/) + x ) 4 x Example 3. Find x x dx. Solution. We could try x = sin(θ) as above, and in all the other problems in this section, that is the right kind of thing to do. But not here. This problem is a u-substitution. Always look for an easy u-substitution first! u = x, du = x dx, du = x dx

48 CHAPTER 7. TECHNIQUES OF INTEGRATION 48 x x dx = u du Example 4. Find x x dx = u 3/ 3/ = 3 ( x ) 3/ = 3 ( x ) 3/ Solution. We follow the substitution on the chart x = 3 tan(θ) dx = 3 sec (θ) 9 + x = 3 sec(θ) We apply all of these substitutions to our integral (3 tan(θ)) 3 3 sec(θ)3 sec (θ) dθ = 3 5 tan 3 (θ) sec 3 (θ) dθ Now we follow our rules of thumb for powers of trig functions. u = sec(θ) du = sec(θ) tan(θ) dθ tan (θ) = sec (θ) And so our integral becomes this (3 tan(θ)) 3 3 sec(θ)3 sec (θ) dθ = 3 5 tan 3 (θ) sec 3 (θ) dθ = 3 5 (sec (θ) ) sec (θ) tan(θ) sec(θ) dθ = 3 5 (u )u du = 3 5 u 4 u du ( ) u = u3 3 ( = sec5 (θ) ) 3 sec3 (θ) ( ) 5 ( ) 3 = x 9 + x ( = (9 + x ) 5/ 3 ) 3 3 (9 + x ) 3/ = 5 (9 + x ) 5/ 3(9 + x ) 3/