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Math Help - Trig Integral

  1. #1
    Member VitaX's Avatar
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    Trig Integral

    \int sin^2(2x)cos^3(2x) dx

    \int sin^2(2x)cos^2(2x)cos(2x) dx

    \int sin^2(2x)(1-sin^2(2x))cos(2x) dx

    let u = sin(2x) du = 2cos(2x)dx \rightarrow \frac{1}{2}du = cos(2x)dx

    \frac{1}{2}\int u^2(1-u^2)du = \frac{1}{2} \int u^2 - u^4 du

    \frac{1}{2}\left[\frac{u^3}{3} - \frac{u^5}{5}\right] + C \rightarrow \frac{1}{2}\left[\frac{5u^3 - 3u^5}{15}\right] + C

    =\frac{1}{6}sin^3(2x) - \frac{1}{10}sin^5(2x) + C

    I thought I did everything right but wolfram got a different answer than me. Is what I did here correct?
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  2. #2
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    Hello, VitaX!

    Your work is correct.

    What was wolfram's answer?

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  3. #3
    Member VitaX's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, VitaX!

    Your work is correct.

    What was wolfram's answer?

    \frac{1}{16}sin(2x) - \frac{1}{96}sin(6x) - \frac{1}{160}sin(10x) + C
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  4. #4
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    Wolfram is telling me that both solutions are equal. You can probably get from one solution to the other using a lot of double angle formulas and the like.

    In the meantime, rest assured that your solution is correct.
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