Integration 1/10. Integration. Student Guidance Centre Learning Development Service

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1 Integration / Integration Student Guidance Centre Learning Development Service lds@qub.ac.uk

2 Integration / Contents Introduction. Indefinite Integration Definite Integration Integration by Substitution 4 Integration by Parts 5 4 Integration Leading to Logarithms 6 5 Integrating Partial Fractions 7 6 Integrating Trigonometric Functions 8 6. Trigonomtric substitution Other Integrals to be aware of Introduction Differentiation and integration are two interrelated areas of mathematics and make up the branch of maths broadly called calculus. You may recall that differentiating a function gives the gradient of a curve. In other words, the derivative describes how the gradient of a function f (x) changes with x. Integration can be thought of as the reverse process of differentiation. So if we find the derivative of a function and then integrate the resulting derivative we return back to the original function. Broadly then, the integral is the inverse function of the derivative. You may recall that differentiating a polynomial: y x n f (x) dy dx nxn Bearing in mind that integration is the reverse process of differentiation if we are to integrate f (x) x n : x n dx xn+ (n + ) +C Instead of bringing the power down and then reducing the power by one, in integration we increase the power by one and then divide by the increased power. Note the above integral is a so-called indefinite integral. It is important to note the constant C added to the evaluated integral. This constant must be added when evaluating all indefinite integrals. The reasoning behind why a constant is needed can be thought of in the following way: if integration is the reverse process of differentiation then when we integrate the derivative of a function we return back to the original function. So an unknown constant is added because the original function may have had an additional constant that would be lost in the process of differentiation.. Indefinite Integration Indefinite integration formally involves finding the antiderivatives of a function f. The process of finding the antiderivative is called integration. In other words, the result of finding the integral of a function gives the function that when differentiated gives back the function we are integrating. Example Find: x 7 dx The antiderivative of x 7 is 8 x8. Hence, this means the x 7 dx 8 x8 + C. We can check our final result by differentiating f (x) 8 x8 +C and the derivative f (x) x 7 (differentiating a constant C is equal to ). Thus it is quite easy to check the answer. If you take the derivative of the final result it should equal the original function (i.e. the integrand). Example Find: (4x + 6x 7x + )dx Now finding the integral or the antiderivative: (4x + 6x 7x + )dx x 4 + x 7 x + x +C Example Find: x 5x + x dx x 5 + x dx Rewriting /x as x we can then carry out the integral: x 5 + x dx x 5x + lnx +C Here we re using the rule that x dx ln x +C. Example 4 Find: 4 x 4 x x + + x dx First we rewrite the integrand in terms of fractional powers before we can carry out the integral: 4x / 4x x + + x / dx Thus the evaluated integral then becomes: 4x / 4x x ++x / dx 4 x/ / 4x ln x +x + x5/ 5/ 8 x/ + 4x ln x + x x5/ ) ) ) x 9 dx x / dx x 4 dx e 4x dx

3 Integration /. Definite Integration Also integration gives the area beneath the curve. Definite integration involves finding the area under a curve where the integration has limits. For example, to integrate the function y x between x and x would be written as: x dx Then carrying out the integration: [ ] x dx x We then substitute in the upper and lower limits and then take the difference: [ ] [ ] [ ] x dx x 7 6 This then represents the area beneath the curve of the function y x, between the lines x and x. Example Find the area between the curve y x(x ) and the coordinates x and x 5. First it is a good idea to sketch the curve to get an idea of the region bounded by x and x 5. If we set the function equal to : y x(x ) we can find the x-coordinates where the function intercepts the x-axis. Thus, the function intercepts the x-axis at x and x. A sketch of the curve is shown below with the two bounded areas A and B: A ydx (x x)dx [ x [ 7 x [ 9 7 ] ( 9) ] [] ] [ ] ( ) 4 Note that in all cases when the area is bounded below the x-axis the area is always negative. So we take the absolute value of the found area. Now we need to find the area B between the curve and the x-axis between the coordinates x and x 5: 5 B ydx 5 (x x)dx [ ] x 5 x [ 5 ( 5) ] [ ] 7 ( 9) So the total area beneath the curve is Evaluate the following integrals: Figure. Pot of the function y x(x ) with the areas A and B delineated. The area B above the axis is positive and the area A below the axis is negative. In general all areas bounded above the axis are positive and all areas bounded below the axis are negative. First we find the area A, between the curve and the x-axis coordinates x and x. As the area is totally below the x-axis we expect the calculated area to be negative: ) ) ) π/ x dx x dx e x dx sin(x) dx

4 Integration 4/. Integration by Substitution Similar to differentiating using substitution, integration by substitution involves simplifying an integral by setting it equal to another variable name. For example, it is difficult to complete the integral: (x + )e x +x+ dx However, by setting u x + x + and then differentiating, du/dx x +. We can then rewrite the integral as: (x + )e u dx but now since we have the integral in terms of u we need to change the variable over which we are integrating over from x to u. Using the derivative of u, we can then rearrange the derivative and write dx du/(x + ). The integral we can then write as: (x + )e u dx (x + )e u du (x + ) So the integral just becomes: e u du e u +C Now we can substitute back for u in the integral and we have evaluated the integral: e u du e u +C e x +x+ +C So now we have fully evaluated the integral. We were able to evaluate the integral by simplifying the original integral and rearranging the derivative of u and substituting for dx. It must be stressed that integration by substitution is a skill that gets easier with practice. By doing more and more integrals it becomes easier to spot which part of the integral should be set equal to u. Sometimes the integral is complex and it is difficult to spot what to substitute in the integral and you may need to try a variety of substitutions until you are left with an integral you are able to evaluate. Example (x + 5)(x + 5x) 7 dx Then setting u x + 5x and this means that du/dx x + 5. Then, substituting u into the integral we are left with: (x + 5)u 7 dx and then replacing dx with du/(x + 5). The integral becomes: u 7 du 8 u8 +C 8 (x + 5x) 8 +C Example (7x + 9)dx We choose u such that u 7x + 9 and then differentiating with respect to x the derivative becomes du/dx 7. Then substituting in for u and dx: 7 u du Carrying out the integral and substituting back for u in terms of x: u 7 du 8 u8 +C 8 (x + 5x) 8 +C Example sin( x) dx x We then choose u such that u x and then differentiating with respect to x the derivative becomes du/dx x. Then rearranging for dx: dx xdu Then, substituting for u and dx in the original integral: sin(u)du cos(u) +C cos( x) +C Where we have substituted back for u in the evaluated integral to put the integral back in terms of x. Integrate the following by making the appropriate substitution: ) (x ) 7 dx ) ) 5) sin(7x )dx x + x dx x dx xsin(x )dx

5 Integration 5/. Integration by Parts As discussed in the previous section, differentiation and integration are closely linked. When we differentiate a function to find the gradient, by integrating the derivative we return to the original function. An important and commonly used rule in differentiation is the product rule. Recall that the product rule of differentiation allows one to differentiate a product of two functions of x, which we usually write as u(x) and v(x): dy(x) v(x) du(x) + u(x) dv(x) dx dx dx y (x) v(x)u (x) + u(x)v (x) We can use the product rule of differentiation to derive the equivalent rule for integrating a product of two functions of x, called integration by parts. If we integrate both sides of Equation we get: uvdx which we can write as: uvdx v du dx dx + v du dx dx uv u dv d(uv) dx dx dx dx () u dv dx () dx Therefore, if we are required to integrate the product of two functions u and v we use the formula derived above. The best way to get familiar with this formula is to demonstrate its use by example. Consider the integral: xe x dx v du dx dx uv u dv dx dx Here we set v x and du/dx e x so this means dv/dx and u e x. Thus, we now have everything we need to use the integration by parts by formula. By substituting into the integration by parts formula: xe x dx xe x dx xe x e x dx xe x e x +C In this example it may seem arbitrary what we chose for v and du/dx. However, it is important to pick carefully in order that it is possible to do the integral. Similar to SOHCAHTOA used to remember the definitions of sin, cos and tan, we can use LATE to correctly decide which term we should differentiate and integrate in the integration by parts formula. In the LATE pneumonic: Logarithms, Algebra, Trigonometry and Exponentials. To find out what we should differentiate in the product we begin at the beginning of LATE with L A T E working through the word until we find one of the two functions in the product to be integrated. The first function we encounter as we move to the right is the function we differentiate (i.e. set equal to v in the formula) and likewise to decide which function to integrate (i.e. set equal to du/dx) we work from the end of LATE. Thus, for the example we just went through we find out what to differentiate by working through L A T E. So, as we move to the right, we find Algebra and as we move to the left we find Exponentials. Thus we differentiate the algebraic term and integrate the exponential term-exactly what was done in the example. Example lnxdx Here it is not obvious we are integrating a product of two functions but we can write lnx as lnx. Then, with use of LATE, we differentiate the lnx function and integrate. Therefore, v ln x and du/dx. This means: dv dx x and u x Now we can substitute into the integration by parts formulae (Equation ) and carry out the final integration: Example xsinxdx lnxdx xlnx x dx xlnx x +C x Using the same ideas as before to decide on the terms to differentiate and integrate, we set v x and du/dx sinx. Carrying out the integration and differentiation: dv and u cosx dx Substituting in the appropriate terms into the integration by parts formulae and carrying out the integration: xsinxdx x( cosx) ( cosx)dx xcosx + cosxdx xcosx + sinx +C Example xcosxdx Similar to Example, here we set v x and du/dx cosx. Differentiating and integrating:

6 Integration 6/ dv dx and u sinx Therefore: xcosxdx xsinx Then carrying out the final integration: xsinx sinxdx sinxdx xsinx + cosx +C 9 Note, in all the worked examples above it was only required to use the integration by parts formula once to evaluate the integral. However, there are cases where you may have to use the integration by parts formula twice to completely evaluate the integral i.e. on the first use of the integration by parts formula you are left with an integral that you still cannot evaluate and so need to use the integration by parts formula again. Based on the previous examples: Integrate the following by integrating by parts twice: x e x dx Answer: x e x 9 xex + 7 ex +C Integrate the following by using the integration by parts formula. In example 5, you will have to integrate by using the integration by parts formula twice: ) xlnxdx ) lnx x 5 dx ) xe 6x dx 5) (x + 5)cos x 5 dx (lnx) dx 4. Integration Leading to Logarithms If the integrand is in a particular form evaluating the integral leads to a logarithm function: f (x) dx ln x +C f (x) i.e. when the numerator is the differential of the denominator we simply take the natural logarithm of the denominator. However, it is not always the case that the numerator is the exact differential of the denominator. In these cases the integrand may have to be manipulated to arrive at the form f (x)/ f (x). For example, in a simple case we may have the integral: x dx However, note that the numerator is not the exact differential of the denominator. This means we need to multiply the numerator by and the denominator by in order not to change the overall integral. This can be written as: x dx ln x +C Note, we haven t changed the overall integral and we have taken the logarithm of the denominator. Example tan(x) dx First we begin by rewriting tan(x) sin(x)/cos(x) and then to get the integrand in the form f (x)/ f (x), sin(x) needs to be multiplied by so we must divide by to leave the integrand unchanged: Example sin(x) dx ln(cosx) +C cos(x) e x e x dx The denominator e x when differentiated leads to e x, very similar to the numerator of e x. To manipulate the integrand such that the numerator is the exact differential of the denominator we need to multiply and divide by. The integral then becomes: e x e x dx

7 Integration 7/ Now we have the integral in the form where the numerator is the exact differential of the denominator. This means we can take the logarithm of the denominator. Evaluating the integral: Example e x e x dx ln(ex ) +C xcosx + sinx dx xsinx First of all we need to think what we would obtain if we are to differentiate the denominator. Using the product rule of differentiation setting u x and v sinx this means du/dx and dv/dx cosx. Thus the differential is: dy sinx + xcosx dx Thus, using the product rule to differentiate the denominator we find that the numerator is the exact differential of the denominator. Thus we can now go about evaluating the integral: xcosx + sinx dx ln(xsinx) +C xsinx Integrate the following using the rule that the integral of a quotient, where the numerator is the derivative of the denominator, is equal to the logarithm of the denominator: ) cotxdx ) + x dx 5e 5x ) dx e5x x + x dx 5) xln(x) dx 5. Integrating Partial Fractions Using the ideas discussed in Section 4, we can integrate more complicated functions. For example, suppose we want to integrate: x 5 (4x )(x + ) dx we cannot integrate this function using any of the techniques we have learned so far. To evaluate an integrand of this form first we need to write the integrand in terms of partial fractions: x 5 (4x )(x + ) A (4x ) + B (x + ) Then we need to solve for both A and B. First we multiply both sides of the equation by (4x )(x + ) and we are left with: x 5 A(x + ) + B(4x ) To solve for A and B we substitute values of x that eliminates either A or B. To solve first for A we substitute x /4: 4 5 A( 4 + ) 9 9A 4 A Then to solve for B we substitute x : 9 9B B Thus, now we can rewrite the integrand in terms of partial fractions: x 5 (4x )(x + ) dx Integrating separately then: and dx ln x + +C (x + ) (4x ) dx (4x ) + (x + ) dx 4 (4x ) dx ln 4x +C Note how in the second case we had to rearrange the integrand to get it in the form f (x)/ f (x) by multiplying and dividing by. Thus, our final answer is: x 5 (4x )(x + ) dx ln x + ln 4x +C

8 Integration 8/ Integrate the following by expressing the integrand in terms of partial fractions and then integrate using the rule f (x)/ f (x)dx ln x +C : ) ) ) x(x + ) dx (x + )(x + ) dx x (x + )(x dx x + (x )(x + 7) dx 6. Integrating Trigonometric Functions It is quite common to come across integrating trigonometric functions. If you remember how to differentiate sinx, cosx and tanx then it is quite simple to work out how to integrate trigonometric functions: y sinx; dy dx cosx y cosx; dy dx sinx cosx sinx +C sinx cosx +C y tanx; dy dx sec x sec x tanx +C So, if we remember the derivatives of the trig functions we can easily work out the integrals. However, you must be very careful not to confuse minus signs. For example, the integral of sinx is cosx and the integral of sinx is cosx. Thus, take care to keep track of minus signs when finding the integral from the known derivatives. However, integrating the simple trig functions aren t the only integrals you will come across. You will come across multiples of trigonometric functions, trig functions raised to a power, the addition of trig functions etc. Most you will not be able to integrate without making a substitution to break it down into an integrateable form: + tan A sec A + cot A csc A Example cos xdx Firstly, we need to use a trigonometric identity to substitute for cos x. The identity we will use in this example will be: cosa cos A Substituting then for cos x we are left with the integral: ( + cosx)dx Now we can integrate as normal. Evaluating the integral: ( + cosx)dx x + sinx +C 4 Example sin(x) cos(x)dx We need to substitute for the integrand in order to carry out the integral. The identity we will use will be: sinacosb sin(a + B) + sin(a B) Substituting for A B x we are left with: sin(x) cos(x) sin(4x) Then, substituting for the integrand in the original integral we are left with: sin(4x)dx cos(4x) +C 8 Example cos(5x) cos(x)dx sinacosb sin(a + B) + sin(a B) cosacosb cos(a B) + cos(a + B) sinasinb cos(a B) cos(a B) sin A + cos B cosa cos A sin A cosa cos A cosa sin A sina sinacosa We need to substitute for the integrand in order to carry out the integral. The identity we will use will be: cosacosb cos(a B) + cos(a + B) Substituting then for A 5x and B x we are left with: cos(5x) cos(x) cos(x) + cos(8x) Then, substituting for the integrand in the original integral we are left with: cos(x) + cos(8x)dx sin(x) + sin(8x) +C 8

9 Integration 9/ More difficult examples can involve rearranging the integrand into a form where a trigonometric substitution can be made and also it may be necessary to integrate by substitution. The following example is a difficult example that demonstrates both techniques: Example 4 sin 4 (x)cos (x)dx First we rewrite the integral as: sin 4 (x)(cos (x) cos(x))dx Now we can substitute for cos (x) using the identity that cos x sin x. The integral then becomes: (sin 4 xcosx sin 6 xcosx)dx Now at this point we can make a substitution by writing u sinx and so du/dx cosx. So substituting for u and replacing dx du/cosx we eliminate the factor of cosx and the integral becomes: (u 4 u 6 )du u5 5 u7 7 +C 6. Trigonomtric substitution If we are asked to find the integral: + x dx Now if we set x tanθ. But you might ask, why would we choose to set x tanθ. This is because the denominator reduces to + tan θ which we know from the trigonometric identity is equal to sec θ (sec θ + tan θ). Now we need to change the variable of the integral from dx to dθ by differentiating x tanθ which gives dx/dθ sec θ. Thus, dx sec θdθ. Now, rewriting the integral: sec θ sec θ dθ dθ θ +C tan x +C This is an important standard result that can be generalised: a + x dx a tan x a +C This is a standard result that is worth being aware of and is useful to be able to derive or look up where necessary. 6. Other Integrals to be aware of Suppose we are asked to find: a x dx We base our choice of the substitution in this case by noticing that the denominator a x is similar in form to the trigonometric identity sin A cos A. Thus to get it into a form that matches the denominator of the integral (a a sin A a cos A). So we try the substitution x asinθ, meaning x a sin θ then dx/dθ acosθ and dx acosθdθ. The integral then becomes: a x dx acosθ dθ a cos θ acosθ dθ acosθ dθ θ +C acosθ dθ a a sin θ Now we substitute for θ by rearranging our substitution x asinθ for θ itself. Thus, θ sin a x. So our integral: a x dx sin x a +C Using a trigonometric substitution where indicated compute the following integrals: x ) dx 6 x Substitute x 4sinθ ) + 4x dx Substitute x tanθ ) sin xcos xdx Hint: use the double angle formula sin x cosx and cos x +cosx sin 4 x cos xdx Hint: use the double angle formula twice to evaluate the integral 5) π/ π/6 cos5x cosxdx Hint: use the product formulae to evaluate the integral

10 Integration / Indefinite Integration ) x +C ) x/ +C ) x +C e4x 4 +C Definite Integration ) ) 6 ) e.95 Answers Integration by Substitution ) 6 (x )7 +C ) cos( 7x) +C 7 ) (x + ) ln( x) +C 5) 4 cos(x ) +C Integration Leading to Logarithms ) ln sinx +C ) ln + x +C ) ln e 5x 5x +C 4 ln + x +C 5) ln ln x +C Integrating Partial Fractions ) x(x + ) dx x x + dx ln x ln x + +C ) ln x + ln x + +C ) ln x ln x 4 +C 5 8 ln x ln x + 7 +C Integrating Trigonometric Functions ) x 6 x + 8sin x 4 +C ) tan x +C ) 4 sinxcos x + 8 cosxsinx + 8 x +C 6 sin xcos x 8 sinxcos x +C + 6 cosxsinx + 6 x +C 5) 8.65 Integration by Parts ) x x ln x 4 +C ) 4ln x + 6x 4 +C ) 6 e6x (6x ) +C 5((x + 5)sin(x/5) + 5cos(x/5)) +C 5)x(ln x lnx + ) +C