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Thread: Intergrating sin^2 x cos^2 x

  1. #1
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    Intergrating sin^2 x cos^2 x

    This question has got me stumped;

    Show that $\displaystyle \sin x \cos x = \frac{1}{2}\sin 2x$, and hence or otherwise find the exact value of $\displaystyle \int_0^\frac{\pi}{8} \sin^2x\cos^2x$.

    The first half is easy, it's the integral that's giving me problems.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    here is what you do

    Quote Originally Posted by Flay View Post
    This question has got me stumped;

    Show that $\displaystyle \sin x \cos x = \frac{1}{2}\sin 2x$, and hence or otherwise find the exact value of $\displaystyle \int_0^\frac{\pi}{8} \sin^2x\cos^2x$.

    The first half is easy, it's the integral that's giving me problems.
    rewrite $\displaystyle \int_0^{\frac{\pi}{8}}sin(x)^2\cdot{cos(x)^2}dx$ as $\displaystyle \int_0^{\frac{\pi}{8}}\bigg(\frac{1-cos(2x)}{2}\bigg)\cdot\bigg(\frac{1+cos(2x)}{2}\bi gg)$...and when you run into the obvious problem you will just look how I took care of it in the integral and apply the same concept except now you have to have $\displaystyle cos(2(2x))$
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    rewrite $\displaystyle \int_0^{\frac{\pi}{8}}sin(x)^2\cdot{cos(x)^2}dx$ as $\displaystyle \int_0^{\frac{\pi}{8}}\bigg(\frac{1-cos(2x)}{2}\bigg)\cdot\bigg(\frac{1+cos(2x)}{2}\bi gg)$...and when you run into the obvious problem you will just look how I took care of it in the integral and apply the same concept except now you have to have $\displaystyle cos(2(2x))$
    $\displaystyle \cos (2A) = \cos^2 A - \sin^2 A = 2 \cos^2 A - 1$. Therefore:

    $\displaystyle \cos^2 A = \frac{1}{2} (\cos (2A) + 1)$.

    You'll need this to deal with the aforementioned obvious problem.
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