# [SOLVED] Integral, checking a result

• Jul 20th 2008, 02:12 PM
arbolis
[SOLVED] Integral, checking a result
I hope this time I didn't make an error. I also need your help for a possible little simplification.
$\int \sqrt{1-x^2}dx$ I call this integral $\Psi \Lambda \Xi$ for the sake of beautiful. I made a trig-sub : Let $x=\sin (\theta), dx=\cos (\theta )d\theta$.
$\Rightarrow \Psi \Lambda \Xi$ $=\int \sqrt{1-\sin ^2 (x)}\cos (\theta)d\theta=\int \sqrt{\cos ^2(\theta)}\cos (\theta)d\theta=\int \cos^2(\theta)d\theta$.
Now integration by parts : let $u'=\cos (\theta)$, $u=\sin (\theta)$, $v=\cos (\theta)$, $v'=-\sin (\theta)$ so our beautiful $\Psi \Lambda \Xi$ is equal to $\sin (\theta)\cos (\theta)+\int \sin^2(\theta)d\theta=\sin (\theta)\cos(\theta)+\int 1-\cos^2 (\theta)d\theta=$ $\sin(\theta)\cos(\theta)+\theta-\int \cos^2(\theta)d\theta \Leftrightarrow 2\int \cos^2(\theta)d\theta= \sin(\theta)\cos(\theta)+\theta \Leftrightarrow \int \cos^2(\theta)d\theta=$ $\frac{\sin(\theta)\cos(\theta)+\theta}{2}+C$. Now I backsub and I get that $\Psi \Lambda \Xi = \frac{x\cos[\arcsin(x)]+\arcsin(x)}{2}+C$. Is my result correct? If no, where did I do errors? And can I simplify more this $\cos[\arcsin(x)]$? Thanks in advance!
• Jul 20th 2008, 02:25 PM
Jhevon
Quote:

Originally Posted by arbolis
I hope this time I didn't make an error. I also need your help for a possible little simplification.
$\int \sqrt{1-x^2}dx$ I call this integral $\Psi \Lambda \Xi$ for the sake of beautiful. I made a trig-sub : Let $x=\sin (\theta), dx=\cos (\theta )d\theta$.
$\Rightarrow \Psi \Lambda \Xi$ $=\int \sqrt{1-\sin ^2 (x)}\cos (\theta)d\theta=\int \sqrt{\cos ^2(\theta)}\cos (\theta)d\theta=\int \cos^2(\theta)d\theta$.
Now integration by parts : let $u'=\cos (\theta)$, $u=\sin (\theta)$, $v=\cos (\theta)$, $v'=-\sin (\theta)$ so our beautiful $\Psi \Lambda \Xi$ is equal to $\sin (\theta)\cos (\theta)+\int \sin^2(\theta)d\theta=\sin (\theta)\cos(\theta)+\int 1-\cos^2 (\theta)d\theta=$ $\sin(\theta)\cos(\theta)+\theta-\int \cos^2(\theta)d\theta \Leftrightarrow 2\int \cos^2(\theta)d\theta= \sin(\theta)\cos(\theta)+\theta \Leftrightarrow \int \cos^2(\theta)d\theta=$ $\frac{\sin(\theta)\cos(\theta)+\theta}{2}+C$. Now I backsub and I get that $\Psi \Lambda \Xi = \frac{x\cos[\arcsin(x)]+\arcsin(x)}{2}+C$. Is my result correct? If no, where did I do errors? And can I simplify more this $\cos[\arcsin(x)]$? Thanks in advance!

overkill.

just recall the identity that $\cos^2 \theta = \frac {1 + \cos 2 \theta}2$ and integrate that. which is a lot easier (and remember to back-substitute when done and write your answer in terms of x)
• Jul 20th 2008, 02:33 PM
arbolis
Quote:

overkill.
Does that mean I'm right?
Quote:

just recall the identity that http://www.mathhelpforum.com/math-he...7f7cdc45-1.gif and integrate that. which is a lot easier (and remember to back-substitute when done and write your answer in terms of x)
I got that $\Psi \Lambda \Xi=\frac{\arcsin (x)}{2}+\frac{\sin(2\arcsin(x))}{4}+C$. Is that good?
• Jul 20th 2008, 02:37 PM
wingless
$\cos \arcsin x = \sqrt{1-x^2}$

Jhevon's solution is the easiest and most common one. It's also possible to solve this without using any trigonometric substitution.

$\int \sqrt{1-x^2}~dx = \int \frac{1-x^2}{\sqrt{1-x^2}}~dx =$ $\arcsin x - \int \frac{x^2}{\sqrt{1-x^2}}~dx = \arcsin x + x\sqrt{1-x^2} -\int \sqrt{1-x^2}~dx$

$\int \sqrt{1-x^2}~dx = \arcsin x + x\sqrt{1-x^2} -\int \sqrt{1-x^2}~dx$

$\int \sqrt{1-x^2}~dx = \frac{\arcsin x + x\sqrt{1-x^2}}{2}$