Example. Evaluate. 3x 2 4 x dx.

Transcription

1 3x 2 4 x dx. Solution: We need a new technique to integrate this function. Notice that if we let u x 3 + 4, and we compute the differential du of u, we get: du 3x 2 dx Going back to our integral, notice that we can eliminate every x that appears in the integral and write it completely in terms of u using both the definition of u and its differential. 3x ( ) 2 4 1/4 ( ) x dx x x 2 dx u 1/4 du In this process, we took an integral that looked difficult and with a substitution we were able to rewrite the integral into a simpler looking one that we can do. Evaluating this gives: 3x 2 4 x dx u 1/4 du Using u x and du 3x 2 dx. 4u5/4 + C 5 Using the power rule for integrals. 4(x3 + 4) 5/4 + C 5 Converting back to x s. AMAT 217 (University of Calgary) Fall / 24

2 Substitution Rule: If u g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f (g(x))g (x) dx f (u) du. Proof: This is proved by differentiating the chain rule. For the function composition F g, the chain rule states (F (g(x))) F (g(x))g (x). Taking antiderivatives (and switching sides) gives: F (g(x))g (x) dx F (g(x)) + C. Now, by setting u g(x) we get: F (g(x))g (x) dx F (u) + C F (u) du. Finally, writing F f, we have: f (g(x))g (x) dx f (u) du. AMAT 217 (University of Calgary) Fall / 24

3 Notes: You can think of the Substitution Rule as being the opposite of the Chain Rule. The Substitution Rule allows us to turn a difficult looking integral into one that is easier to compute. Tip: Choose a substitution u so that its derivate appears in the integral (up to a constant). A general strategy to follow is as follows. Substitution Rule Strategy 1 Choose a possible u u(x), try one that appears inside a function. 2 Calculate du u (x) dx. 3 Either replace u (x) dx by du, or replace dx by 4 Write the rest of the integrand in terms of u. 5 Integrate. 6 Rewrite the result back in terms of x. du u, and cancel. (x) AMAT 217 (University of Calgary) Fall / 24

4 x sin(x 2 ) dx Solution: Since x 2 appears inside of a function we can try the substitution u x 2. Taking the derivative gives: du dx 2x du 2x dx du 2 x dx Since an x dx appears in the integrand, this substitution should work: x sin(x 2 ) dx sin(x 2 ) x dx Rearranging the integrand du sin(u) Using u x 2 and du 2 2 x dx 1 sin(u) du Pulling the constant out in front cos(u) + C Since integral of sin x dx cos x + C. 1 2 cos(x2 ) + C Convert to x using u x 2. AMAT 217 (University of Calgary) Fall / 24

5 the following integral: cos( x) x dx. We use the substitution: Then the derivative is: Solving for dx gives: The integral becomes: u x. du 1 2 x dx. dx cos( x) x dx 2 x du 2 cos u x cos u du 2 x du 2 sin u + C 2 sin( x) + C AMAT 217 (University of Calgary) Fall / 24

6 : 2x 3 x dx. Solution: It makes sense to let: u x Taking derivatives gives: Making this substitution gives: 2x 3 x dx du 2x dx. dx du 2x 2x 3 u du 2x x 2 u du Issue: Our integrals can t have two variables in them (this is not allowed until Calculus III). Usually this means we chose our u incorrectly. But in this case: x 2 u du u x x 2 u 1. (u 1) u du u 3/2 u 1/2 du By expanding products! 2 5 u5/2 2 3 u3/2 + C 2 5 (x2 + 1) 5/2 2 3 (x2 + 1) 3/2 + C AMAT 217 (University of Calgary) Fall / 24

7 sin 3 x cos x dx Solution: We can try u sin x or u cos x (note: both will actually work for this example). Using u sin x and du cos x dx, that is, dx du cos x : sin 3 x cos x dx u 3 cos x u 3 du du cos x Using the substitution Canceling (writing integral in terms of u s) u4 4 + C Since sin4 x 4 x n dx xn+1 + C, n 1 n C Replacing u back in terms of x AMAT 217 (University of Calgary) Fall / 24

8 Products of Sine and Cosine We use the following guidelines to determine a suitable substitution when dealing with products of sine and cosine: Products of Sine and Cosine When evaluating sin m x cos n x dx: 1 The power of sine is odd (m odd): (a) Use u cos x and du sin x dx. (b) Cancel one sin x from the dx replacement to be left with an even number of sines. (c) Use sin 2 x 1 cos 2 x ( 1 u 2 ) to replace the leftover sines. 2 The power of cosine is odd (n odd): (a) Use u sin x and du cos x dx. (b) Cancel one cos x from the dx replacement to be left with an even number of cosines. (c) Use cos 2 x 1 sin 2 x ( 1 u 2 ) to replace the leftover cosines. 3 The power of both sine and cosine are odd (m and n are odd): Use either 1 or 2 (both will work). 4 The power of both sine and cosine are even (m and n are even): Use cos 2 x 1 2 (1 + cos(2x)) and/or sin2 x 1 (1 cos(2x)) to reduce to a form that can 2 be integrated. AMAT 217 (University of Calgary) Fall / 24

9 sin 6 x cos 5 x dx Solution: Power of cos is odd, therefore use u sin x and du cos x dx, that is, dx du cos x : sin 6 x cos 5 x dx u 6 cos 5 x du cos x u 6 ( cos 2 x ) 2 du u 6 (1 sin 2 x) 2 du u 6 (1 u 2 ) 2 du u 6 2u 8 + u 10 du u7 7 2u9 9 + u C Since sin7 x 7 2 sin9 x 9 + sin11 x 11 Using the substitution Canceling a cos x and rewriting cos 4 x Using trig identity cos 2 x 1 sin 2 x Writing integral in terms of u s Expand and collect like terms x n dx xn+1 + C, n 1 n C Replacing u back in terms of x AMAT 217 (University of Calgary) Fall / 24

10 cos 3 x dx Solution: Power of cos is odd, therefore use u sin x and du cos x dx, that is, dx du cos x. This may seem strange at first since we don t have sin x in the question, but it does work! cos 3 x dx cos 3 x cos 2 x du du cos x Using the substitution Canceling a cos x (1 sin 2 x) du Using trig identity cos 2 x 1 sin 2 x (1 u 2 ) du Writing integral in terms of u s u u3 3 + C Since sin x sin3 x 3 x n dx xn+1 + C, n 1 n C Replacing u back in terms of x AMAT 217 (University of Calgary) Fall / 24

11 sin 2 x cos 2 x dx Solution: Power of sine and cosine are both even, thus we use the formulas: cos 2 x 1 (1 + cos(2x)) 2 sin2 x 1 (1 cos(2x)) 2 Rewriting the integrand we have: sin 2 x cos 2 x 1 2 (1 + cos(2x)) 1 (1 cos(2x)) 2 1 (1 + cos(2x)) (1 cos(2x)) 4 1 ( 1 cos 2 (2x) ) 4 Now we have: sin 2 x cos 2 x dx 1 (1 cos 2 (2x)) dx Using trig identities dx 1 cos 2 (2x) dx Splitting up integral 4 4??? AMAT 217 (University of Calgary) Fall / 24

12 Solution (CONTINUED): sin 2 x cos 2 x dx x cos 2 (2x) dx For the second integral, the power of sine and cosine are both even. (Although sine does not appear in the integral, you can think of it as sine to the power of zero and zero is an even number). Thus we use formula: cos 2 θ 1 (1 + cos(2θ)) 2 sin 2 x cos 2 x dx x 4 1 (1 + cos(4x)) dx Using the formula with θ 2x 8 x x 8 sin(4x) 32 ( x + sin(4x) ) + C Use 4 + C Since x/4 x/8 x/8 cos(ax + b) dx 1 sin(ax + b) + C a AMAT 217 (University of Calgary) Fall / 24

13 An example with three answers? Example sin x cos x dx Solution 1: We use the double angle formula sin(2x) 2 sin x cos x. sin x cos x dx 1 sin(2x) dx Replacing sin x cos x by 2 sin 2x 2 This is a standard substitution with u 2x and du 2dx, that is, dx du 2 : sin x cos x dx 1 2 sin(u) du 2 Using the substitution 1 sin u du Pulling out the constant ( cos u) + C Since sin x dx cos x + C 1 cos(2x) + C Replacing u back in terms of x 4 AMAT 217 (University of Calgary) Fall / 24

14 An example with three answers? Example sin x cos x dx Solution 2: Solution 3: We use the substitution u sin x so that du cos x dx. sin x cos x dx u du u2 2 + C Since 1 2 sin2 x + C Using our substitution x n dx xn+1 + C, n 1 n + 1 We use the substitution u cos x so that du sin x dx. sin x cos x dx u du Using our substitution u2 2 + C Since x n dx xn+1 + C, n 1 n cos2 x + C AMAT 217 (University of Calgary) Fall / 24

15 Solution 1 gave an answer of 1 4 cos(2x) + C Solution 2 gave an answer of 1 2 sin2 x + C Solution 3 gave an answer of 1 2 cos2 x + C Question: Which one is correct? Answer: All of them! Calculus I: Two functions that have the same derivative differ by a constant. In each solution the +C is misleading and is actually a different constant for each answer: Answer 1: 1 4 cos(2x)+c 1 1 Answer 2: 2 sin2 x +C 2 Answer 3: 1 2 cos2 x +C 3 Plotting the three functions without constants confirms that they have the same derivative: Algebraically you can also show each of the three functions differ by at most a constant. For example, Answer 2 and Answer 3 are equal when C 3 C : Answer sin2 x + C 2 1 ( ) 1 cos 2 x + C 2 1 ( 2 2 cos2 x + C ) cos2 x + C 3 Answer 3 AMAT 217 (University of Calgary) Fall / 24

16 We next look at integrals with secant and tangent. Some we already know how to do: sec 2 x dx tan x + C sec x tan x dx sec x + C We can also integrate tan x quite easily. Example tan x dx Note that tan x sin x cos x and let u cos x, so that du sin x dx, i.e., dx du sin x : sin x tan x dx dx Rewriting tan x cos x sin x du Using the substitution u sin x 1 u du Cancelling and pulling the 1 out 1 ln u + C Since dx ln x + C x ln cos x + C Replacing u back in terms of x ln sec x + C Using log properties and sec x 1/ cos x AMAT 217 (University of Calgary) Fall / 24

17 Integrating secant Example sec x dx This is a tough integral. You won t need to know how to evaluate this integral for this course. Weird Trick: We multiply the top and bottom by sec x + tan x and use u sec x + tan x. Then, du (sec x tan x + sec 2 x)dx. sec x dx sec x(sec x + tan x) dx sec x + tan x sec 2 x + sec x tan x dx sec x + tan x 1 u du ln u + C Since Doing the trick. Expanding the top. Using the substitution. 1 dx ln x + C x ln sec x + tan x + C Replacing u back in terms of x AMAT 217 (University of Calgary) Fall / 24

18 Products of Secant and Tangent We use the following guidelines to determine a suitable substitution when dealing with products of secant and tangent: Products of Secant and Tangent When evaluating sec m x tan n x dx: 1 The power of secant is even (m even): (a) Use u tan x and du sec 2 x dx. (b) Cancel sec 2 x from the dx replacement to be left with an even number of secants. (c) Use sec 2 x 1 + tan 2 x ( 1 + u 2 ) to replace the leftover secants. 2 The power of tangent is odd (n odd): (a) Use u sec x and du sec x tan x dx. (b) Cancel one sec x and one tan x from the dx replacement. The number of remaining tangents is even. (c) Use tan 2 x sec 2 x 1 ( u 2 1) to replace the leftover tangents. 3 m is even or n is odd: Use either 1 or 2 (both will work). 4 The power of secant is odd and the power of tangent is even: No guidelines. Remember that sec x dx and sec 3 x dx are in integral tables. AMAT 217 (University of Calgary) Fall / 24

19 sec 6 x tan 6 x dx Power of secant is even. Use u tan x, so that du sec 2 x dx, i.e., dx du sec 2 x : sec 6 x tan 6 x dx sec 6 x (u 6 du ) sec 2 Using the substitution x sec 4 x(u 6 ) du Cancelling (sec 2 x) 2 (u 6 ) du Rewriting sec 4 x (1 + tan 2 x) 2 (u 6 ) du Using sec 2 x 1 + tan 2 x (1 + u 2 ) 2 (u 6 ) du Using the substitution u tan x (u 6 + 2u 8 + u 10 ) du Expanding. u u9 9 + u C Since x n dx xn+1 + C, n 1 n + 1 tan7 x tan9 x 9 + tan11 x 11 + C Replacing u back in terms of x AMAT 217 (University of Calgary) Fall / 24

20 sec 5 x tan x dx The power of tangent is odd. Use u sec x, so that du sec x tan x dx, i.e., dx du sec x tan x. Remember: Before we replace sec x by u, we need to save one to cancel!! sec 5 x tan x dx sec 5 x tan x sec 4 x du u 4 du du sec x tan x Substituting dx first Cancelling u5 5 + C Since sec5 x 5 Now substituting u sec x x n dx xn+1 + C, n 1 n C Replacing u back in terms of x AMAT 217 (University of Calgary) Fall / 24

21 sec 5 x tan 3 x dx Power of tan is odd. Use u sec x, so that du sec x tan x dx, i.e., dx Remember: Before we replace sec x by u, we need to save one to cancel!! sec 5 x tan 3 x dx sec 5 x tan 3 du x Substituting dx first sec x tan x sec 4 x tan 2 x du Cancelling u 4 tan 2 x du Now substituting u sec x u 4 (sec 2 x 1) du Using tan 2 x sec 2 x 1 u 4 (u 2 1) du Using the substitution (u 6 u 4 ) du Expanding u7 7 u5 5 + C Since sec7 x 7 sec5 x 5 du sec x tan x. x n dx xn+1 + C, n 1 n C Replacing u back in terms of x AMAT 217 (University of Calgary) Fall / 24

22 Extra Problems: Example sec 2 x tan 6 x dx The power of secant is even. We use u tan x, so that du sec 2 x dx, i.e., dx du sec 2 x : sec 2 x tan 6 x dx sec 2 x (u 6 ) du sec 2 x u 6 du Cancelling u7 7 + C Since tan7 x 7 Using the substitution x n dx xn+1 + C, n 1 n C Replacing u back in terms of x AMAT 217 (University of Calgary) Fall / 24

23 sec 4 x tan 6 x dx The power of secant is even. We use u tan x, so that du sec 2 x dx, i.e., dx du sec 2 x : sec 4 x tan 6 x dx sec 4 x (u 6 du ) sec 2 Using the substitution x sec 2 x(u 6 ) du Cancelling (1 + tan 2 x)(u 6 ) du Using sec 2 x 1 + tan 2 x (1 + u 2 )(u 6 ) du Using the substitution u tan x (u 6 + u 8 ) du Expanding. u7 7 + u9 9 + C Since tan7 x 7 + tan9 x 9 x n dx xn+1 + C, n 1 n C Replacing u back in terms of x AMAT 217 (University of Calgary) Fall / 24

24 sec x tan 2 x dx The power of secant is odd and the power of tangent is even. The guidelines don t help us in this scenario. We first use the formula: tan 2 x sec 2 x 1 sec x tan 2 x dx sec x(sec 2 x 1) dx (sec 3 x sec x) dx Trig identity Expanding 1 (sec x tan x + ln sec x + tan x ) ln sec x + tan x + C Formulas sec x tan x + 1 ln sec x + tan x ln sec x + tan x + C Expand sec x tan x 1 ln sec x + tan x + C Simplify 2 AMAT 217 (University of Calgary) Fall / 24