How to integrate this function? $\int_{-1}^1 \frac{1}{\sqrt{1-x^2}(1+x^2)}\,dx$

Question

How to find definite integral $$\int_{-1}^1 \frac{1}{\sqrt{1-x^2}(1+x^2)}\,dx$$

using complex intergral?

And if $$ f(z) = \frac{1}{\sqrt{1-z^2}(1+z^2)}\,$$

There are simple poles at $$ z = i , z = -i $$ so I calculated residue at those poles, $$ \lim_{z \to i} (z-i)f(z) = \frac{-1}{2\sqrt2} $$ $$ \lim_{z \to -i} (z+i)f(z) = \frac{1}{2\sqrt2} $$

but it seems like I'm wrong, can you guys help me?

btw the answer is $$ \frac{\pi}{\sqrt2} $$

Surely this integral diverges? Did you mean $\sqrt{1+x^2}$ instead of $\sqrt{1-x^2}$? — Ty., Oct 03 '20 at 12:52
oh I made a mistake! thank you, it is from -1 to +1. I confused this with another problem I was solving — thisisyoung__, Oct 03 '20 at 13:30

Z Ahmed · Answer 1 · 2020-10-03T14:07:42.273

1

Let $x=\tan t \implies dx= \sec^2 t dt$, then $$I=2\int_{0}^{1} \frac{1}{\sqrt{1-x^2}(1+x^2)} dx= 2\int_{0}^{\pi/4} \frac{\cos t}{\sqrt{1-2 \sin^2 t}} dt =2\frac{1}{\sqrt{2}}\int_{0}^{1} \frac{dz}{\sqrt{1-z^2}}=2\frac{1}{\sqrt{2}}\sin^{-1}z|_{0}^{1}=\frac{\pi}{\sqrt{2}}.$$ Lastly, we have use $z=\sqrt{2} \sin t$.

edited Oct 03 '20 at 14:07

answered Oct 03 '20 at 13:52

Z Ahmed

43,235

There is a mistake in typo,,, – A learner Oct 03 '20 at 14:02

How to integrate this function? $\int_{-1}^1 \frac{1}{\sqrt{1-x^2}(1+x^2)}\,dx$

1 Answers1