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080129quiz2solution

# 080129quiz2solution - − cos θ 1 3 cos 3 θ C = − √ 1...

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MATH 126 QUIZ #2 Solution 01/31/08 Mathias Knape Last Name : Circle discussion section First Name: 11 am 12 pm 1 pm Note: In order to receive full credit, please show all your work (conceptual steps, deFnitions you use, . ..) 1. Evaluate the integrals (a) Z x sin xdx We use integration by parts to evaluate this integral. Let u = x and dv =s in xdx , then du = dx and v = cos x .Thu s Z x sin xdx = x cos x + Z cos xdx = x cos x +sin x + C. (b) Z tan 3 xdx Z tan 3 xdx = Z tan 2 x tan xdx = Z tan x (sec 2 x 1) dx = Z tan x sec 2 xdx Z tan xdx = 1 2 tan 2 x Z sin x cos x dx = 1 2 tan 2 x +ln | cos x |

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(c) Z x 3 1 x 2 dx Let x =s in θ for π 2 θ π 2 .Then dx =cos θdθ and Z x 3 1 x 2 dx = Z sin 3 θ p 1 sin 2 θ cos θdθ = Z sin 3 θ cos θ cos θdθ = Z (1 cos 2 θ )sin θdθ
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Unformatted text preview: − cos θ + 1 3 cos 3 θ + C = − √ 1 − x 2 + 1 3 (1 − x 2 ) 3 2 + C. 2. First make a substitution and then use integration by parts to evaluate the integral. (a) Z 4 1 e √ x dx We substitute w = √ x , then dx = 2 wdw . Thus Z 4 1 e √ x dx = Z 2 1 e w 2 wdw. Next we use integration by parts. Let u = 2 w and dv = e w dw , then du = 2 dw and v = e w . Therefore Z 2 1 e w 2 wdw = e w 2 w | 2 1 − Z 2 1 e w 2 dw = 4 e 2 − 2 e − 2 e 2 + 2 e = 2 e 2 ....
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080129quiz2solution - − cos θ 1 3 cos 3 θ C = − √ 1...

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