Find $\int \frac{\sqrt{1-x^2}}{1+x^2}\hspace{1mm}dx$ Find $$\int \dfrac{\sqrt{1-x^2}}{1+x^2}\hspace{1mm}dx$$
Any hints! I will do the work, just give me a clue 
 A: Hints:
$1)$ put $x = \sin u$
$2)$  $$\frac{\cos^2 u}{1 + \sin^2 u} = \frac{1}{1 + 2 \tan^2 u} $$
$3)$ put $t = \tan u$
A: You're looking for hints only, so try
Step 1: sub $x=\sin\theta$
Step 2: then sub $t=\tan\theta$
Then you're there...
A: Notice, $$\int\frac{\sqrt{1-x^2}}{1+x^2}dx$$ Let, $x=\sin\alpha\implies dx=\cos\alpha d\alpha$   $$\int \frac{\sqrt{1-\sin^2\alpha}}{1+\sin^2\alpha}\cos\alpha d\alpha $$ $$=\int\frac{\cos^2\alpha}{1+\sin^2\alpha}d\alpha$$  $$=\int\frac{\sec^4\alpha\cos^2\alpha}{\sec^4\alpha(1+\sin^2\alpha)}d\alpha$$
$$=\int\frac{\sec^2\alpha}{(1+\tan^2\alpha)(\sec^2\alpha+\sec^2\alpha\sin^2\alpha)}d\alpha$$
$$=\int\frac{\sec^2\alpha}{(1+\tan^2\alpha)(1+\tan^2\alpha+\tan^2\alpha)}d\alpha$$
$$=\int\frac{\sec^2\alpha}{(1+\tan^2\alpha)(1+2\tan^2\alpha)}d\alpha$$ Let, $\tan\alpha=t\implies \sec^2\alpha d\alpha=dt$
$$=\int\frac{dt}{(1+t^2)(1+2t^2)}dt$$  $$=\int\left(\frac{2}{1+2t^2}-\frac{1}{1+t^2}\right)dt$$  $$=2\int\frac{dt}{1+(t\sqrt 2)^2}-\int\frac{dt}{1+t^2}dt$$ 
$$=\sqrt 2\tan^{-1}\left(t\sqrt 2\right)-\tan^{-1}\left(t\right)+c$$ 
Setting $t=\tan \alpha$
$$=\sqrt 2\tan^{-1}\left(\sqrt 2\tan\alpha\right)-\tan^{-1}\left(\tan\alpha\right)+c$$ 
Setting $\tan \alpha=\frac{\sin\alpha}{\cos \alpha}=\frac{x}{\sqrt{1-x^2}}$
$$=\sqrt 2\tan^{-1}\left(\frac{x\sqrt 2}{\sqrt{1-x^2}}\right)-\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)+c$$ 
