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!


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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  • $\begingroup$ you want to rewrite this part as a function of $f'(y)$ $\endgroup$ – tired Jun 8 '15 at 15:42
  • $\begingroup$ @BetterWorld you want to know the motivation? Well, since I can't compute $f'(y)$ directly, I need an equation in $f'(y)$. If you look at the fourth line of the solution you posted, there is an $f'(y)$ present in the calculation. This comes from $2y \int_0^{\pi/2} \tan^2 x/(\tan^2 x + y^2)\, dx$. The trick is to add and subtract $y^2$ to $\tan^2 x$ to get $$\frac{\tan^2 x}{\tan^2 x + y^2} = 1 - \frac{y^2}{\tan^2 x + y^2}$$ Then integrate. $\endgroup$ – kobe Jun 8 '15 at 16:16

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