# Thread: Stuck @ integral :S

1. ## Stuck @ integral :S

hello
I really stucked at this one

$\displaystyle \int arctan\left( \sqrt{\frac{1-x}{1+x}} \right) dx$

2. Let $\displaystyle I=\int arctan\left( \sqrt{\frac{1-x}{1+x}} \right) dx$

Substitute $\displaystyle x=cos(\theta)$

$\displaystyle \implies \frac{1-x}{1+x}=\frac{1-cos(\theta)}{1+cos(\theta)}$

$\displaystyle =\frac{2 \, sin^2(\frac{\theta}{2})}{2 \, cos^2(\frac{\theta}{2})}$

Hence,
$\displaystyle \sqrt{\frac{1-x}{1+x}}=tan\left(\frac{\theta}{2}\right)$

So, $\displaystyle I=-\int arctan\left( tan\left(\frac{\theta}{2}\right) \right) \, sin(\theta) \, d\theta$

$\displaystyle =\frac{-1}{2}\int \theta \, sin(\theta) \, d\theta$

A quick integration by parts will finish it ..

3. Perfect!!

thanks

4. Wait. how did you know substituting cosx for x will solve the problem?

5. By my mind.

6. Thanks.

7. Originally Posted by lilaziz1
Wait. how did you know substituting cosx for x will solve the problem?
He's just very, very good at this!

8. Originally Posted by Miss
$\displaystyle \int arctan\left( \sqrt{\frac{1-x}{1+x}} \right) dx$
$\displaystyle \displaystyle\int\frac{dt}{(1+t^2)^2}=\frac12\left (\frac t{t^2+1}+\arctan t\right)+k_1,$ now in your integral substitute $\displaystyle x=\dfrac{1-t^{2}}{1+t^{2}}$ so it becomes $\displaystyle -4\displaystyle\int{\frac{t\arctan t}{\left( t^{2}+1 \right)^{2}}\,dt}=-4\int{\left( -\frac{1}{2\left( 1+t^{2} \right)} \right)'\arctan t\,dt}$ which is equal to $\displaystyle \displaystyle\frac{2\arctan t}{1+t^2}-2\int\frac{dt}{(1+t^2)^2}$ and problem solved.