nice improper integral asked by A.c. Kesharwani in the calculus form group

Exercise:

Show that , $\int_{0}^{\pi }{\frac{n}{{{n}^{2}}+{{\sin }^{2}}x}dx}=\frac{\pi }{\sqrt{1+{{n}^{2}}}}\,\,\,\,,\,\,\,\,n>0$

Solution: Let ${{f}_{n}}\left( x \right)=\frac{n}{{{n}^{2}}+{{\sin }^{2}}x}$

Observe that , ${{f}_{n}}\left( \pi -x \right)=\frac{n}{{{n}^{2}}+{{\sin }^{2}}\left( \pi -x \right)}=\frac{n}{{{n}^{2}}+{{\sin }^{2}}x}={{f}_{n}}\left( x \right)$

So ${{f}_{n}}$ is a periodic function of period $\pi $

So $\int_{0}^{\pi }{\frac{n}{{{n}^{2}}+{{\sin }^{2}}x}dx}=2\int_{0}^{\frac{\pi }{2}}{\frac{n}{{{n}^{2}}+{{\sin }^{2}}x}dx}=2n\int_{0}^{\frac{\pi }{2}}{\frac{dx}{{{n}^{2}}+{{\sin }^{2}}x}}$

We have ${{n}^{2}}+{{\sin }^{2}}x={{n}^{2}}\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)+{{\sin }^{2}}x=\left( 1+{{n}^{2}} \right){{\sin }^{2}}x+{{n}^{2}}{{\cos }^{2}}x$

So $\int_{0}^{\frac{\pi }{2}}{\frac{dx}{{{n}^{2}}+{{\sin }^{2}}x}}=\int_{0}^{\frac{\pi }{2}}{\frac{dx}{\left( 1+{{n}^{2}} \right){{\sin }^{2}}x+{{n}^{2}}{{\cos }^{2}}x}}$

Now look to $\left( 1+{{n}^{2}} \right){{\sin }^{2}}x+{{n}^{2}}{{\cos }^{2}}x={{\cos }^{2}}x\left[ \left( 1+{{n}^{2}} \right){{\tan }^{2}}x+{{n}^{2}} \right]$

So $\int_{0}^{\frac{\pi }{2}}{\frac{dx}{{{n}^{2}}+{{\sin }^{2}}x}=\int_{0}^{\frac{\pi }{2}}{\frac{dx}{{{\cos }^{2}}x\left[ \left( 1+{{n}^{2}} \right){{\tan }^{2}}x+{{n}^{2}} \right]}}}$

                       $=\int_{0}^{\frac{\pi }{2}}{\frac{{{\sec }^{2}}x}{\left( 1+{{n}^{2}} \right){{\tan }^{2}}x+{{n}^{2}}}dx}=\int_{0}^{\frac{\pi }{2}}{\frac{{{\sec }^{2}}x}{{{\left( \sqrt{1+{{n}^{2}}}\tan x \right)}^{2}}+{{n}^{2}}}dx}$

Let $u=\tan x\Leftrightarrow du={{\sec }^{2}}xdx$

So $\int_{0}^{\frac{\pi }{2}}{\frac{{{\sec }^{2}}x}{{{\left( \sqrt{1+{{n}^{2}}}\tan x \right)}^{2}}+{{n}^{2}}}dx}=\int_{0}^{\frac{\pi }{2}}{\frac{du}{{{\left( \sqrt{1+{{n}^{2}}}u \right)}^{2}}+{{n}^{2}}}}$

Let \(u=\frac{n}{\sqrt{1+{{n}^{2}}}}w\Leftrightarrow du=\frac{n}{\sqrt{1+{{n}^{2}}}}dw\)

So $\int{\frac{du}{{{\left( \sqrt{1+{{n}^{2}}}u \right)}^{2}}+{{n}^{2}}}}=\frac{n}{\sqrt{1+{{n}^{2}}}}\int{\frac{dw}{{{\left( nw \right)}^{2}}+{{n}^{2}}}=\frac{1}{n\sqrt{1+{{n}^{2}}}}\int{\frac{dw}{{{w}^{2}}+1}}}$

                              $=\frac{1}{n\sqrt{1+{{n}^{2}}}}\arctan \left( \frac{\sqrt{1+{{n}^{2}}}}{n}u \right)+c$

Hence $\int_{0}^{\frac{\pi }{2}}{\frac{{{\sec }^{2}}x}{{{\left( \sqrt{1+{{n}^{2}}}\tan x \right)}^{2}}+{{n}^{2}}}dx=\frac{1}{n\sqrt{1+{{n}^{2}}}}\arctan \left( \frac{\sqrt{1+{{n}^{2}}}}{n}\tan x \right)_{0}^{\frac{\pi }{2}}}$

Thus $\int_{0}^{\pi }{\frac{n}{{{n}^{2}}+{{\sin }^{2}}x}dx=\frac{2}{\sqrt{1+{{n}^{2}}}}\left( \arctan \left( \infty  \right)-0 \right)=\frac{\pi }{\sqrt{1+{{n}^{2}}}}}$

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the idea of solution credit to Slimane Zerguine