Inti....!

$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + (asinx)^{2}}$

$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{sec^{2}xdx}{1 + tan^{2}x + (atanx)^{2}}$

$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{sec^{2}xdx}{(1 + a^{2})tan^{2}x +1}$

$tanx = t$

$\displaystyle \int_{0}^{\infty} \frac{tdt}{(1 + a^{2})t^{2} +1}$

$= \frac{1}{\sqrt{1 + a^{2}}}[_{0}^{\infty}tan^{-1}t]$

$= \frac{1}{\sqrt{1 + a^{2}}} \frac{\pi}{2}$

Simple

Or else you can do is, put a=0 then integrate it :D

This one is not too difficult. Look we see that $\sin x$ has even power so the best way is to divide the numerator and denominator by $\cos ^{2} x$ . Dividing we get:- $\int_{0}^{\dfrac{\pi}{2}} \dfrac{\sec^{2}x}{sec^{2}x+a^{2}\tan^{2}x}dx$ Now put $tan x =t$ $\sec^{2} x dx = dt$ We will also change the limit

When $x=0, t=0$

When $x=\dfrac{\pi}{2} , t=\infty$

Now the integral becomes $\int_{0}^{\infty} \dfrac{1}{1+t^{2}+a^{2}t^{2}}dt$ $=\dfrac{1}{1+a^{2}}\int_{0}^{\infty} \dfrac{1}{\dfrac{1}{1+a^{2}}+t^{2}}dt$ We can solve this by standard methods of substitution of $\tan^{-1}x$ . Hence the integral is solved as $=\Big[\dfrac{1}{\sqrt{1+a^{2}}}\tan^{-1}\dfrac{t}{\sqrt{1+a^{2}}}\Big]_{0}^{\infty}$ Solving the limts we get the answer as $=\dfrac{1}{\sqrt{1+a^{2}}} \Big[\dfrac{\pi}{2}-1\Big]$ $=\dfrac{\pi}{2\sqrt{1+a^{2}}}$