Too Clever?

Relevant wiki: Integration of Trigonometric Functions - Intermediate

$\displaystyle \large f(n) = \int_{0}^{\pi /2}\dfrac{dx}{1+n\sin ^2 x} = \int_{0}^{\pi/2} \dfrac{dx}{n+1-n\cos^2 x} =$

$\displaystyle \large \int_{0}^{\pi/2}\dfrac{dx}{n+1-\frac{n}{1+\tan^2 x}} = \int_{0}^{\pi/2}\dfrac{\sec^2 x}{1+(n+1)\tan^2 x}dx$

Now substituting $u=\tan x \implies du = \sec ^2x dx$ and changing the limits of integration,

$\displaystyle \large \implies f(n)=\int_{0}^{\infty} \dfrac{du}{1+(n+1)u^2} = \dfrac{\arctan(x\sqrt{n+1})}{\sqrt{n+1}}|_{0}^{\infty}$

$\displaystyle \large \implies \boxed{f(n)=\dfrac{\pi}{2\sqrt{n+1}}}$

Alternatively, we could have used $\sin^2 x = \frac{1}{1+\cot^2 x}$ , but I find dealing with $tangent$ function 'easier' than $co-tangent$ .