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Math Help - [SOLVED] Help with trig reduction formula

  1. #1
    Junior Member eddie2042's Avatar
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    [SOLVED] Help with trig reduction formula

    OK so when we're evaluating integrals of the form

    \int(sin^n x)dx

    we know reduction formulas from the past that make it simpler to evaluate them which is

    \int \sin^n x dx = - \frac{\ \sin^{n-1} x \cos x}{n} + \frac{\ n-1}{n} \int \sin^{n-2} x dx

    however my problem is when the term inside the  sin isn't just an x

    In my case, it's

    \int sin ^2 (5x)dx

    What do i do? I'm pretty sure it changes something because I'm off by a factor. My final answer is

     -\frac{1}{4}sin(10x) + \frac{x}{2} + C

    However the correct answer is

    -\frac{1}{20}sin(10x) + \frac{x}{2} + C

    I'm off by a factor of 5. Coincidence?
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  2. #2
    Junior Member eddie2042's Avatar
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    Never mind i solved it :P
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