# Very last step of a trigonometric integral substitution?

• Apr 11th 2011, 05:29 PM
DannyMath
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)
• Apr 11th 2011, 05:52 PM
topsquark
Quote:

Originally Posted by DannyMath
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, $\displaystyle sin(2u) = 2~sin(u)~cos(u)$

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

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

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

-Dan
• Apr 11th 2011, 06:00 PM
DannyMath
Ah yes one of the most important identities. Makes it much easier, thank you!
• Apr 12th 2011, 05:04 AM
TheCoffeeMachine
An alternative method to Wolfram's would be integration by parts:

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

Spoiler:
\displaystyle \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}

\displaystyle \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}
• Apr 12th 2011, 06:15 AM
DannyMath
Cool, thanks!