The goal of today is to determine what u-substitution to use for trigonometric integrals. The most common substitutions are the following:

Transcription

1 Trigonometric Integrals The goal of today is to determine what u-substitution to use for trigonometric integrals. The most common substitutions are the following: Substitution u sinx u cosx u tanx u secx derivative du cosxdx du sin xdx du sec 2 xdx du secx tan xdx Quite often, we need to use a trigonometric formula to turn the integral from x s to u s. Trig Identities sin 2 x +cos 2 x 1 tan 2 x +1sec 2 x cos 2 x 1 2 (1 + cos(2x)) sin2 x 1 2 (1 cos(2x)) cos(2x) cos 2 x sin 2 x sin(2x) 2sinx cos x It is quicker to skip substitution rule when integrating sin(ax + b) orcos(ax + b): cos(ax + b) dx 1 sin(ax + b)+c sin(ax + b) dx 1 cos(ax + b)+c a a Math 267 (University of Calgary) Fall / 17

2 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 xdx: 1 The power of sine is odd (m odd): (a) Use u cosx and du sin xdx. (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 )toreplacetheleftoversines. 2 The power of cosine is odd (n odd): (a) Use u sinx and du cosxdx. (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 )toreplacetheleftovercosines. 3 Both m and n are odd: Use either 1 or 2 (both will work). 4 Both m and n are even: Use cos 2 x 1 2 (1 + cos(2x)) and/or sin2 x 1 2 (1 be integrated. cos(2x)) to reduce to a form that can Math 267 (University of Calgary) Fall / 17

3 Evaluate sin 3 x cos xdx Solution: We can try u sinx or u cosx (both will work but one is easier than the other). Using u sinx and du cosxdx,thatis, dx du cos x : sin 3 x cos x dx u 3 cos x du cos x Using the substitution u 3 du Canceling (writing integral in terms of u s) u4 4 + C Since x n dx xn+1 n +1 + C, n 6 1 sin4 x 4 + C Replacing u back in terms of x Math 267 (University of Calgary) Fall / 17

4 Evaluate cos 2 xdx. Solution: cos 2 xdx (1 + cos(2x)) dx Using trig identity cos 2 x 1 (1 + cos(2x)) 2 x + sin(2x) + C Since cos(ax + b) dx 1 sin(ax + b)+c 2 a x 2 + sin(2x) 4 + C Math 267 (University of Calgary) Fall / 17

5 Evaluate sin 6 x cos 5 xdx Solution: Power of cos is odd, thereforeuseu sinx and du cosxdx,thatis, dx du cos x : sin 6 x cos 5 x dx u7 7 sin7 x 7 u 6 cos 5 x du cos x Using the substitution u 6 cos 2 x 2 du Canceling a cos x and rewriting cos 4 x u 6 (1 sin 2 x) 2 du Using trig identity cos 2 x 1 sin 2 x u 6 (1 u 2 ) 2 du Writing integral in terms of u s u 6 2u 8 + u 10 du Expand and collect like terms 2u u C Since x n dx xn+1 n +1 + C, n 6 1 2sin 9 x + sin11 x + C Replacing u back in terms of x 9 11 Math 267 (University of Calgary) Fall / 17

6 Evaluate cos 3 xdx Solution: Power of cos is odd, thereforeuseu sinx and du cosxdx,thatis, 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 u sinx cos 3 x cos 2 xdu 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 C Since sin 3 x 3 x n dx xn+1 n +1 + C Replacing u back in terms of x + C, n 6 1 Math 267 (University of Calgary) Fall / 17

7 We next look at integrals with secant and tangent. Some we already know how to do: sec 2 xdxtanx + C sec x tan xdxsecx + C We can also integrate tan x quite easily. Evaluate tan xdx Note that tan x sin x cos x and let u cosx, sothatdu sin xdx,i.e., dx du sin x : sin x tan xdx dx Rewriting tan x cos x sin x du Using the substitution u sin x 1 du Cancelling and pulling the 1out u 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 Math 267 (University of Calgary) Fall / 17

8 Evaluate tan 2 xdx Note that tan 2 x sec 2 x 1. tan 2 xdx sec 2 1xdx Rewriting tan x tanx x + C Since sec 2 xdxtanx + C Math 267 (University of Calgary) Fall / 17

9 Integrating secant Evaluate sec xdx This is a tough integral. You won t need to know how to evaluate this integral. It ll be on the formula sheet. Weird Trick: We multiply the top and bottom by sec x +tanx and use u secx +tanx. Then, du (secx tan x +sec 2 x)dx. sec xdx sec x(sec x +tanx) dx sec x +tanx 1 u du sec 2 x +secx tan x sec x +tanx dx ln u + C Since Doing the trick. Expanding the top. Using the substitution. 1 x dx ln x + C ln sec x +tanx + C Replacing u back in terms of x Math 267 (University of Calgary) Fall / 17

10 Evaluate sec 3 xdx This is an even tougher integral which will also be on the formula sheet. We use integration by parts: u dv u v v du. 2 It turns out that the best choice for u and dv is u secx and dv sec 2 xdx. 8 >< u secx dv sec 2 xdx derivative & >: du secx tan xdx v tanx sec 3 xdx secx tan x secx tan x secx tan x secx tan x secx tan x 9 > integral >; sec x tan 2 xdx Integration by Parts formula sec x(sec 2 x 1) dx Using tan 2 x sec 2 x 1 (sec 3 sec x) dx Expand sec 3 dx + sec xdx Sum/di erence rule sec 3 dx +ln sec x +tanx + C 0 We know R sec xdx from last example sec 3 xdx secx tan x +ln sec x +tanx + C 0 Combine common integral sec 3 xdx 1 (sec x tan x +ln sec x +tanx ) +C Divide by two 2 Math 267 (University of Calgary) Fall / 17

11 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 xdx: 1 The power of secant is even (m even): (a) Use u tanx and du sec 2 xdx. (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 )toreplacetheleftoversecants. 2 The power of tangent is odd (n odd): (a) Use u secx and du secx tan xdx. (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 xdx and sec 3 xdx are on the formula sheet. Math 267 (University of Calgary) Fall / 17

12 Evaluate sec 2 x tan 6 xdx The power of secant is even. We use u tanx, sothatdu sec 2 xdx,i.e., dx du sec 2 x : sec 2 x tan 6 xdx sec 2 x (u 6 ) u 6 du du sec 2 x Using the substitution Cancelling u7 7 + C Since tan7 x 7 x n dx xn+1 n +1 + C Replacing u back in terms of x + C, n 6 1 Math 267 (University of Calgary) Fall / 17

13 Evaluate sec 5 x tan 3 xdx Power of tan is odd. Useu secx, sothatdu secx tan xdx,i.e., dx du sec x tan x. Remember: Before we replace sec x by u, weneedtosaveonetocancel!! sec 5 x tan 3 xdx sec 5 x tan 3 du x Substituting dx first sec x tan x sec 4 x tan 2 xdu Cancelling u 4 tan 2 xdu Now substituting u secx 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 u C Since x n dx xn+1 + C, n 6 1 n +1 sec7 x sec 5 x + C Replacing u back in terms of x 7 5 Math 267 (University of Calgary) Fall / 17

14 Evaluate sec x tan 2 xdx 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 xdx sec x(sec 2 x 1) dx Trig identity (sec 3 x sec x) dx Expanding 1 2 (sec x tan x +ln sec x +tanx ) ln sec x +tanx + C Formulas 1 2 sec x tan x ln sec x +tanx ln sec x +tanx + C Expand 1 2 sec x tan x 1 ln sec x +tanx + C Simplify 2 Math 267 (University of Calgary) Fall / 17

15 Multiple Arguments What if we encounter combinations of sin(a 1 x + b 1 )andcos(a 2 x + b 2 )? Multiple Argument Formulas When evaluating sin(a 1 x + b 1 )sin(a 2 x + b 2 ) dx, use: When evaluating sin A sin B 1 [cos(a B) cos(a + B)] 2 cos(a 1 x + b 1 )cos(a 2 x + b 2 ) dx, use: When evaluating cos A cos B 1 2 [cos(a sin(a 1 x + b 1 )cos(a 2 x + b 2 ) dx, use: B)+cos(A + B)] sin A cos B 1 2 [sin(a B)+sin(A + B)] Also recall the following two formulas: cos(ax + b) dx 1 sin(ax + b)+c a sin(ax + b) dx 1 a cos(ax + b)+c Math 267 (University of Calgary) Fall / 17

16 Evaluate sin(3x)cos(12x) dx We use the following formula: sin A cos B 1 2 [sin(a B)+sin(A + B)] In our question, A 3x and B 12x, thus: sin(3x)cos(12x) [sin(3x 12x)+sin(3x +12x)] [sin( 9x)+sin(15x)] 1 2 [ sin(9x)+sin(15x)] sin(3x)cos(12x) dx 1 ( sin(9x)+ sin(15x)) dx Trig identity 2 1 cos(9x) cos(15x) + C By R sin(ax + b) dx formula cos(9x) 18 cos(15x) 30 + C Expanding Math 267 (University of Calgary) Fall / 17

17 Evaluate sin(12x +3)sin(3x +1)dx We use the following formula: sin A sin B 1 2 [cos(a B) cos(a + B)] In our question, A 12x +3andB 3x +1,thus: sin(12x +3)sin(3x +1) [cos((12x +3) (3x +1)) cos((12x +3)+(3x +1))] [cos(9x +2) cos(15x +4)] sin(12x +3)sin(3x +1)dx 1 (cos(9x +2) cos(15x +4))dx Trig identity 2 1 sin(9x +2) sin(15x +4) + C By R cos(ax + b) dx sin(9x +2) sin(15x +4) + C Expanding Math 267 (University of Calgary) Fall / 17