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Math Help - [SOLVED] Trig substitution

  1. #1
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    [SOLVED] Trig substitution

    <br />
\int_0^\frac {\pi}{5} sin^4(5x)~dx<br />

    I know how to solve this problem if it was only a second power, but since it is to the fourth power, I am a little confused.
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  2. #2
    Super Member 11rdc11's Avatar
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    Use half angle identity to break it down
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  3. #3
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    Quote Originally Posted by redman223 View Post
    <br />
\int_0^\frac {\pi}{5} sin^4(5x)~dx<br />

    I know how to solve this problem if it was only a second power, but since it is to the fourth power, I am a little confused.
    <br />
\int_0^\frac {\pi}{5} sin^4(5x)~dx<br />

    =  \int_0^\frac {\pi}{5} \left[ sin^2(5x) \right]^2~dx

    =  \int_0^\frac {\pi}{5} \left[ \frac{1- cos(10x)}{2} \right]^2~dx

    =  \int_0^\frac {\pi}{5} \left[ \frac{1- 2cos(10x) + cos^2(10x)}{4} \right]~dx

    =  \frac {1}{4} \int_0^\frac {\pi}{5} \left[1- 2cos(10x) \right]~dx +  \frac {1}{4} \int_0^\frac {\pi}{5} cos^2(10x) ~dx

    =  \frac {1}{4} \int_0^\frac {\pi}{5} \left[1- 2cos(10x) \right]~dx +  \frac {1}{8} \int_0^\frac {\pi}{5} \left[1+cos(20x) \right] ~dx
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  4. #4
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    So is

    <br />
sin^4(5x) = sin^2(5x)sin^2(5x)<br />

    Edit: Looks like we posted at the same time, I guess it is equal, thanks.
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