# How to evaluate $\int_0^\frac{1}{\sqrt2}\frac{dx}{(1+x^2)\sqrt{1-x^2}}$

$$\int_0^\frac{1}{\sqrt2}\frac{dx}{(1+x^2)\sqrt{1-x^2}}$$

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.

## 3 Answers

hint: $$\frac{1}{1+\sin^2 x}=\frac{\sec^2 x}{1+2\tan^2 x}$$

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 $$\frac{\arctan\sqrt2}{\sqrt2}\approx 0.675510858856$$

• Why was the hint @EHE gave not enough? Why do the calculation for the OP? – JavaMan Apr 1 at 2:43
• @JavaMan Because it was an opportunity to demonstrate something more than the OP asked, namely $I(x;a,b,c)$. If I were a beginning calc. student I would find it very helpful, even interesting, to see such an answer. IMO, a good answer helps you in more ways than $1$, and helps you connect concepts/ideas (and in this case integrals). – clathratus Apr 1 at 3:31
• Of course I found it very useful. Using a second substitution is somewhat challenging task. Thank you for pointing it out! – emil Apr 1 at 15:26
• @emil You are very welcome :) – clathratus Apr 1 at 15:45

Hint

Divide numerator & denominator by $$\sin^2\theta$$

and set $$\cot\theta =u$$

Or divide numerator & denominator by $$\cos^2\theta$$

and set $$\tan\theta=v$$