Question

If $f\left(k\right)=\underset{-1}{\overset{1}{\int }}\frac{\sqrt{1-x^2}}{\sqrt{k+1}-x}dx,$ then the value of $\sum _{k=0}^{99}\frac{f\left(k\right)}{\pi }$ is

Answer 1

$f\left(k\right)=\underset{-1}{\overset{1}{\int }}\frac{\sqrt{1-x^2}}{\sqrt{k+1}-x}dx$
Putting $x\to -x$, we get
$\Rightarrow f\left(k\right)=\underset{-1}{\overset{1}{\int }}\frac{\sqrt{1-x^2}}{\sqrt{k+1}+x}dx$
Adding both, we get
$\Rightarrow f\left(x\right)=\sqrt{k+1}\underset{-1}{\overset{1}{\int }}\frac{\sqrt{1-x^2}}{k+(1-x^2)}dx\Rightarrow f\left(x\right)=2\sqrt{k+1}\underset{0}{\overset{1}{\int }}\frac{\sqrt{1-x^2}}{k+(1-x^2)}dx$

Putting $x=\sin \theta \Rightarrow dx=\cos \theta d\theta$ $f\left(k\right)=2\sqrt{k+1}\underset{0}{\overset{\pi /2}{\int }}\frac{{\cos }^2\theta }{k+{\cos }^2\theta }d\theta =2\sqrt{k+1}\underset{0}{\overset{\pi /2}{\int }}\frac{k+{\cos }^2\theta -k}{k+{\cos }^2\theta }d\theta =2\sqrt{k+1}\left[\underset{0}{\overset{\pi /2}{\int }}d\theta -k\underset{0}{\overset{\pi /2}{\int }}\frac{d\theta }{k+{\cos }^2\theta }\right]=2\sqrt{k+1}[\frac{\pi }{2}-k\underset{0}{\overset{\pi /2}{\int }}\frac{{\sec }^2\theta d\theta }{1+k(1+{\tan }^2\theta )}]$

Putting $\tan \theta =z\Rightarrow {\sec }^2\theta d\theta =dz$
$=2\sqrt{k+1}[\frac{\pi }{2}-k\underset{0}{\overset{\infty }{\int }}\frac{dz}{1+k+k{z}^2}]=2\sqrt{k+1}[\frac{\pi }{2}-\frac{k}{k+1}\underset{0}{\overset{\infty }{\int }}\frac{dz}{1+{\left(\sqrt{\frac{k}{k+1}}z\right)}^2}]=2\sqrt{k+1}[\frac{\pi }{2}-\sqrt{\frac{k}{k+1}}{\left({\tan }^{-1}\left(\sqrt{\frac{k}{k+1}}z\right)\right)}_0^{\infty }]=2\sqrt{k+1}[\frac{\pi }{2}-\frac{\pi }{2}\frac{\sqrt{k}}{\sqrt{k+1}}]\therefore \frac{f\left(k\right)}{\pi }=\sqrt{k+1}-\sqrt{k}$

Now,
$\sum _{k=0}^{99}\frac{f\left(k\right)}{\pi }=\sum _{k=0}^{99}(\sqrt{k+1}-\sqrt{k})=(\sqrt{1}-0)+(\sqrt{2}-\sqrt{1})+\dots +(\sqrt{100}-\sqrt{99})=\sqrt{100}-0=10$