S2 - Math 128: Solution 2 Koray Karabina...

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Math 128: Solution 2 Koray Karabina (kkarabin@uwaterloo.ca) 1. Evaluate the following integrals: (a)[ 10 points ] R sin 4 x cos 3 xdx (b)[ 0 points ] R tan 5 xdx (c)[ 10 points ] R tan x + tan 4 xdx (d)[ 0 points ] R π/ 2 0 sin 2 2 θ dθ (e)[ 10 points ] R cos πx cos 4 πxdx Hint: cos A cos B = 1 2 (cos ( A - B ) + cos ( A + B )) (f)[ 10 points ] R cos x cos 3 (sin x ) dx (g)[ 0 points ] R cot 3 x sin 4 xdx Solution: (a) The integrand is a product of sin and cos functions, where the power of cos is odd. Hence, we can write Z sin 4 x cos 3 xdx = Z sin 4 x cos 2 x cos xdx = Z sin 4 x (1 - sin 2 x ) cos xdx = Z u 4 (1 - u 2 ) du ( u = sin x, du = cos xdx ) = u 5 5 - u 7 7 + C = sin 5 x 5 - sin 7 x 7 + C (b) Z tan 5 dx = Z (tan 2 x ) 2 tan xdx = Z (sec 2 x - 1) 2 tan xdx (tan 2 x = sec 2 x - 1) = Z sec 4 x tan xdx - 2 Z sec 2 x tan xdx + Z tan xdx = I 1 + - 2 I 2 + I 3 , (1) where I 1 = R sec 4 x tan xdx , I 2 = R sec 2 x tan xdx , and I 3 = R tan xdx . Now, the integrand in I 1
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is a product of sec and tan functions, where the power of sec is even. Hence, we can evaluate I 1 as I 1 = Z sec 4 x tan xdx = Z (sec 2 x ) tan x sec 2 xdx = Z (1 + tan 2 x ) tan x sec 2 xdx = Z (1 + u 2 ) udu ( u = tan x, du = sec 2 xdx ) = u 2 2 + u 4 4 + C 1 = tan 2 x 2 + tan 4 x 4 + C 1 Similarly, I 2 = Z tan x sec 2 xdx = Z udu ( u = tan x, du = sec 2 xdx ) = u 2 2 + C 2 = tan 2 x 2 + C 2 Also, I 3 = Z tan xdx = Z sin x cos x dx = - Z du u ( u = cos x, du = - sin x ) = - ln( | u | ) + C 3 = - ln( | cos x | ) + C 3 Finally, we can write (1) as follows: Z tan 5 dx = tan 4 x 4 - tan 2 x 2 - ln( | cos x | ) + C (c) R tan x + tan 4 xdx = I 1 + I 2 , where I 1 = R tan xdx and I
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S2 - Math 128: Solution 2 Koray Karabina...

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