Evaluate : $\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$ Evaluate: $$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$$
First approach : 
$$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4(1-\cos^2x)}$$
$$=\int^{\frac{\pi}{2}}_0 \frac{\cos^2xdx}{4 - 3\cos^2x}$$
$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3}\{\frac{4-3\cos^2x-4}{4 - 3\cos^2x}\}\,dx$$
$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3}\{ 1- \frac{4}{4 - 3\cos^2x}\}\,dx$$
$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3} 1\,dx- \int^{\frac{\pi}{2}}_0 \frac{1}{3} \frac{4\sec^2x}{4\sec^2x - 3}\,dx$$
$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3} 1\,dx- \int^{\frac{\pi}{2}}_0 \frac{1}{3} \frac{4
\sec^2x}{4(1+\tan^2x) - 3}\,dx$$
Now I can easily put $\tan x = t$ and I get $\sec^2x \,dx =dt$
Second approach : 
$$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$$
Dividing numerator and denominator by $\cos^2x$ we get : 
$$=\int^{\frac{\pi}{2}}_0 \frac{dx}{1 +4\tan^2x}$$
$$=\int^{\frac{\pi}{2}}_0 \frac{dx}{(4)\{\frac{1}{4} +\tan^2x\}}$$
$$=\int^{\frac{\pi}{2}}_0 \frac{dx}{(4)\{\{\frac{1}{2}\}^2 +(\tan x)^2\}}$$
Can we apply this formula of integral here : 
$$\int \frac{1}{a^2+x^2}dx = \frac{1}{a}\tan^{-1}\frac{x}{a}$$
I tried but its not working here, I think doing some manipulation we can implement this here.. Please suggest thanks...

You multiply and divide by $\sec^{4}(x)$ and see if it's working. The denominator becomes $$\sec^{2}(x) + 4 \tan^{2}(x) \sec^{2}(x)$$ Now use $1+\tan^{2}(x)=\sec^{2}(x)$.
So 


*

*In numerator you have $\sec^{2}(x)$

*In denominator you have $\sec^{2}(x) \cdot \left(1+4\tan^{2}(x)\right) = (1+\tan^{2}(x))\cdot(1+4\tan^{2}(x))$.

*Now put $t=\tan(x)$ and your integral becomes $$\int_{0}^{\infty} \frac{dt}{(1+t^{2}) \cdot (1+4t^{2})} = \int_{0}^{\infty} \left\{ \frac{1}{1+t^2} -\frac{4}{1+4t^2}\right\} \ dt$$

Oh. I missed a trick here. Note that you had $$\int_{0}^{\pi/2} \frac{dx}{1+4\tan^{2}(x)}$$ Now put $t = \tan(x)$. Then you have $dt = \sec^{2}(x) \ dx $ and so $dx = \frac{dt}{(1+t^{2})}$.
