TITLE: Using differentiation under integral sign to calculate $\int^{\pi/2}_0\frac{\log(1+a\sin\phi)}{\sin\phi}d\phi$
QUESTION [16 upvotes]: I want to calculate the integral $$\int^{\pi/2}_0\frac{\log(1+\sin\phi)}{\sin\phi}d\phi$$ using differentiation with respect to parameter in the integral $$\int^{\pi/2}_0\frac{\log(1+a\sin\phi)}{\sin\phi}d\phi$$
I know that I have to solve from differentiate under the integral and I must use a suitable substitution for integrands involving trigonometric functions but I can't complete the solution. Could you help me?

REPLY [4 votes]: We can use the following Taylor expansion
\begin{eqnarray} 
\ln(1+x)&=&\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}x^n.
\end{eqnarray}
Then
\begin{eqnarray}
\int_0^{\pi/2}\frac{\ln(1+\sin\phi)}{\sin\phi}d\phi&=&\int_0^{\pi/2}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\sin^{n-1}\phi d\phi\\
&=&\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\frac{\sqrt{\pi}\Gamma(\frac{n}{2})}{2\Gamma(\frac{n+1}{2})}\\
&=&\frac{\sqrt{\pi}}{2}\sum_{n=1}^\infty\frac{\Gamma(\frac{2n-1}{2})}{(2n-1)\Gamma(n)}-\frac{\sqrt{\pi}}{4}\sum_{n=1}^\infty\frac{\Gamma(n)}{n\Gamma(\frac{2n+1}{2})}\\
\end{eqnarray}
Let
$$ f(x)=\frac{\sqrt{\pi}}{2}\sum_{n=1}^\infty\frac{\Gamma(\frac{2n-1}{2})}{(2n-1)\Gamma(n)}x^{2n}, g(x)=\frac{\sqrt{\pi}}{4}\sum_{n=1}^\infty\frac{\Gamma(n)}{n\Gamma(\frac{2n+1}{2})}x^{2n}. $$
Then
$$ \left(\frac{f(x)}{x}\right)'=\frac{\sqrt{\pi}}{2}\sum_{n=1}^\infty\frac{\Gamma(\frac{2n-1}{2})}{\Gamma(n)}x^{2n-2}=\frac{\pi}{2\sqrt{1-x^2}} $$
and hence
$$ f(x)=\frac{\pi}{2} x\arcsin x. $$
Also 
$$ g(x)=\frac{1}{2}\arcsin^2x. $$
So $f(1)=\frac{\pi^2}{4}, g(1)=\frac{\pi^2}{8}$ and hence
$$ \int_0^{\pi/2}\frac{\ln(1+\sin\phi)}{\sin\phi}d\phi=\frac{\pi^2}{8}. $$