How do you find the definite integral of cos^4(2x)sin(2x) dx?

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How do you find the definite integral of cos^4(2x)sin(2x) dx?

Please help guide me through this question with steps. I just want to know how to approach this question. I tried using u-substitution with u = 2x but I am unable to find the answer. Thanks for your time.

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Answer 1 · 2017-12-17T23:45:30

$- \frac{1}{5}$ $-1/5$

Explanation:

Remember that the point of u-substitutions is to eliminate complex polynomials or functions. Under this mentality it would make sense that we try to eliminate the trigonometric functions sine and cosine. We will let $u = cos (2 x)$ $u = cos(2x)$ , so that $d u = - 2 sin (2 x) d x$ $du = -2sin(2x) dx$ .

$\int_{1}^{- 1} u^{4} sin (2 x) \cdot \frac{d u}{- 2 sin (2 x)}$ $int_1^(-1) " "u^4 sin(2x) * (du)/(-2sin(2x))$

Notice how the $sin (2 x)$ $sin(2x)$ s cancel out, making our u-subsitution very useful. We are now left with

$\int_{1}^{- 1}$ $int_1^(-1)$ $- \frac{1}{2} u^{4}$ $-1/2u^4$ $d u$ $du$

which yields $- \frac{1}{10} u^{5}$ $-1/10u^5$ evaluated from 1 to -1. Thus, our final answer is $- \frac{1}{5}$ $color(red)(-1/5)$ .

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How do you find the definite integral of cos^4(2x)sin(2x) dx?

Please help guide me through this question with steps. I just want to know how to approach this question. I tried using u-substitution with u = 2x but I am unable to find the answer. Thanks for your time.

1 Answer

Explanation:

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