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

overkill.

just recall the identity that $\displaystyle \cos^2 \theta = \frac {1 + \cos 2 \theta}2$ 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 $\displaystyle \Psi \Lambda \Xi=\frac{\arcsin (x)}{2}+\frac{\sin(2\arcsin(x))}{4}+C$. Is that good?
• Jul 20th 2008, 02:37 PM
wingless
$\displaystyle \cos \arcsin x = \sqrt{1-x^2}$

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

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

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

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