TITLE: How to evaluate $\int_0^\frac{1}{\sqrt2}\frac{dx}{(1+x^2)\sqrt{1-x^2}}$
QUESTION [5 upvotes]: $\int_0^\frac{1}{\sqrt2}\frac{dx}{(1+x^2)\sqrt{1-x^2}}$
I tried using substitute, $x=\sin\theta$
But I ended up with $\int_0^\frac{\pi}{4}\frac{d\theta}{1+\sin^2\theta}$
Is my substitution correct? Please give me a hint to work this out! Thank you very much.

REPLY [3 votes]: We see that 
$$\frac1{1+\sin(x)^2}=\frac{\sec(x)^2}{1+2\tan(x)^2}$$
as was hinted by @E.H.E. We then preform the sub $u=\tan(x)$ to get
$$\int_0^1 \frac{du}{1+2u^2}$$
Then we may in fact compute the integral
$$I(x;a,b,c)=\int\frac{ dx}{ax^2+bx+c}=\int\frac{dx}{a(x+\frac{b}{2a})^2+g}$$
Here $g=c-\frac{b^2}{4a}$. If we assume that $4ac>b^2$, then we may make the substitution $x+\frac{b}{2a}=\sqrt{\frac{g}{a}}\tan u$ which gives 
$$I(x;a,b,c)=\sqrt{\frac{g}{a}}\int\frac{\sec^2u\, du}{g\tan^2u+g}$$
$$I(x;a,b,c)=\frac{u}{\sqrt{ag}}$$
$$I(x;a,b,c)=\frac2{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}}+C$$
And by noting that your integral is given by $I(1;2,0,1)-I(0;2,0,1)$ we have your integral at the value
$$\frac{\arctan\sqrt2}{\sqrt2}\approx 0.675510858856$$