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Math Help - Very last step of a trigonometric integral substitution?

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    Very last step of a trigonometric integral substitution?

    Hi,

    I'm trying to understand the last step shown in this guide to solving this integral (click show steps). I don't know how to simplify that arcsin inside the sin. Any help is appreciated.

    http://www.wolframalpha.com/input/?i=integrate+sqrt(4-x^2)
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    Quote Originally Posted by DannyMath View Post
    Hi,

    I'm trying to understand the last step shown in this guide to solving this integral (click show steps). I don't know how to simplify that arcsin inside the sin. Any help is appreciated.

    http://www.wolframalpha.com/input/?i=integrate+sqrt(4-x^2)
    First, sin(2u) = 2~sin(u)~cos(u)

    so you need to calculate \displaystyle sin \left ( sin^{-1} \left ( \frac{x}{2} \right ) \right ) = \frac{x}{2} (so long as -\pi/2 < x/2 < \pi/2.)

    and \displaystyle cos \left ( sin^{-1} \left ( \frac{x}{2} \right ) \right ).

    Construct a right triangle. Call the base angle \theta, the side across from the angle x and the hypotenuse 2. (This ensures that sin( \theta ) = x/2..) The adjacent side of the triangle is thus \sqrt{4 - x^2}. And this gives us cos( \theta ) = \frac{1}{2}~\sqrt{4 - x^2}.

    -Dan
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    Ah yes one of the most important identities. Makes it much easier, thank you!
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    An alternative method to Wolfram's would be integration by parts:

    \displaystyle \int f'(x) g(x)\;{dx} =  f(x)g(x)-  \int f(x) g'(x)\;{dx} (by parts).

    Spoiler:
    \begin{aligned}\displaystyle I & = \int(x)'\sqrt{a^2-x^2}\;{dx} \\ & = x\sqrt{a^2-x^2}-\int{x}\left(\sqrt{a^2-x^2}\right)'\;{dx}\\ & = x\sqrt{a^2-x^2}+\int \frac{x^2}{\sqrt{a^2-x^2}}\;{dx}\\ & = x\sqrt{a^2-x^2}+\int \frac{a^2-(a^2-x^2)}{\sqrt{a^2-x^2}}\;{dx}\\ & =  x\sqrt{a^2-x^2}+\int \frac{a^2}{\sqrt{a^2-x^2}}\;{dx}-\int \frac{a^2-x^2}{\sqrt{a^2-x^2}}\;{dx}\\ & = x\sqrt{a^2-x^2}+a^2\int \frac{1}{\sqrt{a^2-x^2}}\;{dx}-\int \sqrt{a^2-x^2}\;{dx}\\ & = x\sqrt{a^2-x^2}+a^2\sin^{-1}\frac{x}{a}\;{dx}-I\\ & \end{aligned}

    \begin{aligned}\therefore \space & 2I = x\sqrt{a^2-x^2}+a^2\sin^{-1}\frac{x}{a}\;{dx}+k \\& \therefore I = \frac{1}{2}\left(x\sqrt{a^2-x^2}+a^2\sin^{-1}\frac{x}{a}\right)+k. \end{aligned}
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    Cool, thanks!
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