How to evaluate the definite integral $\int_0^{\pi/2}\ln(2015+\sin^2 x)dx$ 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. 
 A: On the interval $\left[0,\frac{\pi}{2}\right]$ the integrand function is almost constant, since it increases from $\log(2015)$ to $\log(2016)$. In particular, we have:
$$ I=\int_{0}^{\pi/2}\log\left(2015+\sin^2 x\right)\,dx = \frac{\pi}{2}\log(2015)+\int_{0}^{\pi/2}\log\left(1+\frac{\sin^2 x}{2015}\right)\,dx \tag{1}$$
but since:
$$ \log(1+z) = \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\,z^n \tag{2}$$
for any $|z|<1$ and:
$$ \int_{0}^{\pi/2}\sin^{2n}(x)\,dx = \frac{\pi}{2\cdot 4^n}\binom{2n}{n}\tag{3}$$
it follows that:
$$ I = \frac{\pi}{2}\log(2015)+\sum_{n\geq 1}\frac{(-1)^{n+1}\pi}{2n\cdot 8060^n }\binom{2n}{n}.\tag{4} $$
Moreover, since:
$$ \sum_{n\geq 1}\frac{z^n}{n}\binom{2n}{n}=2\log 2-2\log\left(1+\sqrt{1-4z}\right) \tag{5}$$
can be proved by differentiating both sides and exploiting the generating function for Catalan numbers, we have:

$$ I = \frac{\pi}{2}\cdot\log\left(\frac{4031}{4}+6\sqrt{28210}\right).\tag{6}$$

Not by chance, $\frac{4031}{4}+6\sqrt{28210}$ is very close to $2015.5$:
$$ \frac{4031}{4}+6\sqrt{28210} = 2015.4999689903245\ldots $$
A: 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 $a=2015$.
