What is the solution for $\int_{0}^{\pi/2}\frac{\cos^2x}{\cos^2x+4\sin^2x}\,dx$

Question

I used the rule :$$\int_{0}^{a}f(x) = \int_{0}^{a}f(a-x)\,dx$$ And got :$$\int_{0}^{\pi/2}\frac{\sin^2x}{\sin^2x+4\cos^2x}\,dx$$ Then, I added both the question and the above integrand, and I got :$$\int_{0}^{\pi/2}\frac{dx}{5}$$ Solving the above, I finally got the answer as $$\frac{\pi}{20}$$ But, according to my textbook, the answer is $$\frac{\pi}{6}$$ Where did I go wrong?

Answer 1

So what you have done is apparently wrong. The equality you used is true, the problemes lies elsewhere.

In fact when you add the two terms you found :

$$A =\frac{\cos^2x}{\cos^2x+4\sin^2x}$$ $$B =\frac{\sin^2x}{\sin^2x+4\cos^2x}$$

$$\implies A + B = \frac {3 \cos(x)^4 + 3 \sin(x)^4+1}{(3 \sin(x)^2 +1) \cdot (3 \cos(x)^2 + 1)} $$

Here is an attempt to solve the exercice:

$$I = \int_{0}^{\pi/2}\frac{\cos^2x}{\cos^2x+4\sin^2x}dx = \int_{0}^{\pi/2}\frac{1}{1+4\tan^2x}dx $$ we can do this because the interval of integration permits it. Then we change the variable of integration ; the substitution is the following :

$$ u = \tan x$$ we get :

$$I =\int_{0}^{\infty}\frac{1}{(1 +4u^2)(1 + u^2)}du = \int_{0}^{\infty} \frac {4}{3(4u^2 + 1)} - \frac {1}{3(u^2+1)} du $$

from there we can easily conclude.

In fact you just have to know this formula :

$$\int \frac 1 {x^2 + a^2} = \frac 1 {|a|} \arctan \left(\frac x a \right) + C $$ Where $C$ is the constant of integration.

And thus we get :

$$\int_{0}^{\infty} \frac {4}{3(4u^2 + 1)} du = \frac 2 3 \arctan(2u)\bigg\rvert_0^{\infty} = \pi / 3$$ $$\int_{0}^{\infty} \frac {1}{3(u^2+1)} du = \frac 1 3 \arctan u \bigg\rvert_0^{\infty} = \pi/6 $$