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...
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