TITLE: How to evaluate the definite integral $\int_0^{\pi/2}\ln(2015+\sin^2 x)dx$
QUESTION [7 upvotes]: I have to find the value of : 
$$\int_0^{\pi/2}\ln(2015+\sin^2 x)\,dx$$ 
but I have no clue about an effective technique. I tried integration by parts but of no help. I found this on an online site of calculus. I am currently a XIIth grader. Although the integral can be well approximated by trapezoidal and Simpsons rule i want a method to find the exact value.

REPLY [12 votes]: Hint. You may set
$$
I(a):=\int_0^{\pi/2}\ln(a+\sin^2 x)dx, \qquad a\geq0.
$$
Then differentiating under the integral sign with respect to $a$ you get
$$
\begin{align}
I'(a)&=\int_0^{\pi/2}\frac1{a+\sin^2 x}dx\\\\
&=\int_0^{\infty}\frac1{a+\dfrac{t^2}{t^2+1}}\dfrac{dt}{t^2+1}\quad (t=\tan x)\\\\
&=\int_0^{\infty}\frac1{(a+1)t^2+a}dt\\\\
&=\frac{\pi}2\frac{1}{\sqrt{a(a+1)}}\\\\
&=\pi \left.\left(\ln \left(\sqrt{a}+\sqrt{a+1}\right)\right)\right|_a^{'}
\end{align}
$$ Thus
$$
\int_0^{\pi/2}\ln(a+\sin^2 x)dx=\pi \ln \left(\sqrt{a}+\sqrt{a+1}\right)+C
$$ with $C=I(0)=-\pi \ln 2$ (this one is classic) giving

$$
\int_0^{\pi/2}\ln(a+\sin^2 x)dx=\pi \ln \left(\frac{\sqrt{a}+\sqrt{a+1}}2\right), \quad a\geq0.$$

Your initial integral is obtained with $a=2015$.