Question

${\int }_0^1\frac{dx}{(1+x^2)\sqrt{(2+x^2)}}=\frac{\pi }{k}$. Find the value of $k$.

Answer 1

Let $I=\int \frac{1}{(x^2+1)\sqrt{x^2+2}}dx$
Put $x=\sqrt{2}\tan t\Rightarrow dx=\sqrt{2}{\sec }^{2}tdt$
$I=\sqrt{2}\int \frac{sect}{\sqrt{2}(2{\tan }^{2}t+1)}dt=\int \frac{sect}{2{\tan }^{2}t+1}dt$
Multiplying numerator and denominator by ${\cos }^{2}t$, we get
$I=\int \frac{\cos t}{2{\sin }^{2}t+{\cos }^{2}t}dt=\int \frac{\cos t}{{\sin }^{2}t+1}dt$
Now put $u=\sin t\Rightarrow du=\cos tdt$
Therefore
$I=\int \frac{1}{u^2+1}du={\tan }^{-1}u={\tan }^{-1}\sin t$
$={\tan }^{-1}\sin \left({\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)\right)={\tan }^{-1}\left(\frac{x}{\sqrt{x^2+2}}\right)$
Hence
${\int }_0^1\frac{1}{(x^2+1)\sqrt{x^2+2}}dx={\left[{\tan }^{-1}\left(\frac{x}{\sqrt{x^2+2}}\right)\right]}_0^1$
$=\frac{\pi }{6}-0=\frac{\pi }{6}$
$\Rightarrow k=6$