<s>[INST] Integrate $f(y)=\int_{0}^{\frac{\pi}{2}} \ln(y^2 \cos^2x+ \sin^2x) .dx$ This Integral came up while attempting another question: 

$$f(y)=\int_{0}^{\frac{\pi}{2}} \ln(y^2 \cos^2x+ \sin^2x) .dx$$

The suggested solution was as follows:
$$f'(y) = 2y \int_{0}^{\pi/2}\frac{cos^{2}x}{sin^{2}x + y^{2}cos^{2}x}dx$$
$$= 2y \int_{0}^{\pi/2}\frac{dx}{tan^{2}x + y^{2}}$$
$$= 2y \int_{0}^{\pi/2}\frac{sec^{2}x - tan^{2}x }{tan^{2}x + y^{2}}dx$$
$$= 2y . \frac{1}{y} tan^{-1}( \frac{1}{y}) |_{0}^{\infty} -2y\frac{\pi}{2} + y^{2}f'(y)$$
$$f'(y) = \frac{\pi}{1 + y}$$
Unfortunately, I was unable to understand the how to get the last 3 steps. As per my understanding, 
$$f'(y)= 2y \int_{0}^{\pi/2}\frac{sec^{2}x - tan^{2}x }{tan^{2}x + y^{2}}dx$$
Splitting into 2 integrals, 
$$= 2y\int_0^{\pi/2}\dfrac{\sec^2 x}{\tan^2 x + y^2}dx-2y \int_{0}^{\pi/2}\frac{tan^{2}x }{tan^{2}x + y^{2}}dx$$
Substituting $\tan(x)=u$ in the first integral,
$$ 2y\int_0^{\pi/2}\dfrac{\sec^2 x}{\tan^2 x + y^2}dx$$
$$= 2y\int_0^{\infty}\dfrac{du}{u^2 + y^2}$$
$$=2y\times \dfrac{1}{y}\tan^{-1}(\dfrac{u}{y})|_0^\infty$$
However, this does not seem to agree with the given solution. Also, I have no idea as to how to integrate $-2y \int_{0}^{\pi/2}\frac{tan^{2}x }{tan^{2}x + y^{2}}dx$ 
ould somebody please be so kind as to point out my error in computing the first integral and also help me integrate the second integral? Many, many thanks in advance!
 [/INST] Your calculation of $2y\int_0^{\pi/2}\sec^2 x/(\tan^2 x + y^2)\, dx$ is correct so far, and the value is $\pi$. Now
\begin{align}2y\int_0^{\pi/2} \frac{\tan^2 x}{\tan^2 x + y^2}\, dx &= 2y\int_0^{\pi/2} \left(1 - \frac{y^2}{\tan^2 x + y^2}\right)\, dx \\
&= 2y\cdot\frac{\pi}{2}  - y^2\cdot 2y\int_0^{\pi/2} \frac{1}{\tan^2 x + y^2}\, dx\\
&= \pi y - y^2 f'(y),
\end{align}
and therefore
$$2y\int_0^{\pi/2} \frac{\sec^2 x - \tan^2 x}{\tan^2x + y^2}\, dx = \pi - \pi y + y^2f'(y),$$
that is, 
$$f'(y) = \pi - \pi y + y^2 f'(y).$$
Solving for $f'(y)$,
$$f'(y) = \frac{\pi - \pi y}{1 - y^2} =  \frac{\pi(1 - y)}{(1 - y)(1 + y)} = \frac{\pi}{1 + y}.$$
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