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Math Help - How should I approach this problem?

  1. #1
    Member FalconPUNCH!'s Avatar
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    How should I approach this problem?

    We just started learning about Trigonometric Integrals and I was doing pretty good with the odd numbered ones, until I got to the even. I don't really know how to approach this problem.

     \int_{0}^{Pi} sin^{2}t cos^{4}t dt

    Any help would be great.
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  2. #2
    Member FalconPUNCH!'s Avatar
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    Well I think I got the answer and it took me about two pages to get it trail and error.

    Here's how I approached it:

    I substituted Sin^{2}t with  1 - cos^{2}t

     \int_{0}^{Pi} (1-cos^{2}t) (cos^{2}t) (cos^{2}t)

    I used half angle formula to get

     \frac{1}{8} \int_{0}^{Pi} (1- \frac{1}{2}[1+cos^{2}t])(\frac{1}{2}[1+cos^{2}t])(\frac{1}{2}[1+cos^{2}t])

    I expanded everything and got

    \frac{1}{8}\int_{0}^{Pi} 2cos2t + cos^{2}t * cost dt

    I went from there and did a few more steps and got \frac{29Pi}{128}

    Can anyone check if I did it correct?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    Well I think I got the answer and it took me about two pages to get it trail and error.

    Here's how I approached it:

    I substituted Sin^{2}t with  1 - cos^{2}t

     \int_{0}^{Pi} (1-cos^{2}t) (cos^{2}t) (cos^{2}t)

    I used half angle formula to get

     \frac{1}{8} \int_{0}^{Pi} (1- \frac{1}{2}[1+cos^{2}t])(\frac{1}{2}[1+cos^{2}t])(\frac{1}{2}[1+cos^{2}t])

    I expanded everything and got

    \frac{1}{8}\int_{0}^{Pi} 2cos2t + cos^{2}t * cost dt

    I went from there and did a few more steps and got \frac{29Pi}{128}

    Can anyone check if I did it correct?
    that is incorrect. the answer should be \frac {\pi}{16}

    you were pretty much on track about how to start. so you went wrong in the algebra or evaluation somewhere

    use the fact that \sin^2 x = \frac {1 - \cos 2x}2 and \cos^2 x = \frac {1 + \cos 2x}2

    so, \int_0^{\pi} \sin^2 x \cos ^4 x~dx = \frac 18 \int_0^{\pi} (1 - \cos 2x)(1 + \cos 2x)^2~dx = \frac 18 \int_0^{\pi} \sin^2 2x (1 + \cos 2x)~dx

    do a similar manipulation for \sin^2 2x as we did before and continue
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  4. #4
    Member FalconPUNCH!'s Avatar
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    Thanks yeah the page got messy so I might have multiplied incorrectly.
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