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Math Help - Integration with trig functions

  1. #1
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    Integration with trig functions

    1) Integral of the inverse of sinx from 0 to 1/2
    2) Integral of (sinx)^3 * (cosx)^2

    1) Is there an identity I should be looking for when dealing with inverse sine?

    2) Should I be using integration by parts in this?
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  2. #2
    MHF Contributor red_dog's Avatar
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    1) Integration by parts

    2) \int\sin^3x\cos^2xdx=\int(1-\cos^2x)\cos^2x\sin xdx
    Put \cos x=u
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  3. #3
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    Quote Originally Posted by red_dog View Post
    1) Integration by parts

    2) \int\sin^3x\cos^2xdx=\int(1-\cos^2x)\cos^2x\sin xdx
    Put \cos x=u
    should it be just (1-cos^2x) or that over 2? In my textbook it reads:

    sin^2x = (1-cos^2x) / 2
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  4. #4
    MHF Contributor red_dog's Avatar
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    \sin^2x+\cos^2x=1\Rightarrow\sin^2x=1-\cos^2x
    \displaystyle\sin^2x=\frac{1-\cos 2x}{2}
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  5. #5
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    Quote Originally Posted by red_dog View Post
    \sin^2x+\cos^2x=1\Rightarrow\sin^2x=1-\cos^2x
    \displaystyle\sin^2x=\frac{1-\cos 2x}{2}
    thanks.
    for the first one, what do I set dV equal to if i have u = sin x ?
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  6. #6
    MHF Contributor red_dog's Avatar
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    \displaystyle\int_0^{\frac{1}{2}}\arcsin xdx=\int_0^{\frac{1}{2}}x'\arcsin xdx=\left.x\arcsin x\right|_0^{\frac{1}{2}}-\int_0^{\frac{1}{2}}\frac{x}{\sqrt{1-x^2}}dx=
    \displaystyle=\frac{\pi}{12}+\left.\sqrt{1-x^2}\right|_0^{\frac{1}{2}}=\frac{\pi}{12}+\frac{\  sqrt{3}}{2}-1
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  7. #7
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    Quote Originally Posted by xfyz View Post
    1) Integral of the inverse of sinx from 0 to 1/2
    \int\arcsin x\,dx (integration limits omitted).

    Apply integration by parts, so

    u=\arcsin x\implies du=\frac1{\sqrt{1-x^2}}\,dx & dv=dx\implies v=x

    I assume you know integration by parts formula, this yields

    \int\arcsin x\,dx=x\arcsin x-\int\frac x{\sqrt{1-x^2}}\,dx

    For the remaining integral, let's set y=\sqrt{1-x^2}\implies dy=-\frac x{\sqrt{1-x^2}}\,dx

    Therefore \int\arcsin x\,dx=x\arcsin x+\sqrt{1-x^2}+k

    Then we happily can say that

    \int_0^{1/2}\arcsin x\,dx=\left.x\arcsin x+\sqrt{1-x^2}\right|_0^{1/2}, yielding red_dog's answer
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