Integral $\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \ln(1+c\sin x) dx$, where $0I am trying to evaluate the following integral:
$$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \ln(1+c\sin x) dx,$$
where $0<c<1$.
I can't really think of a way to find it so please give me a hint.
 A: Express the integrand as a series:
$$\begin{align}I(c) &= \int_{-\pi/2}^{\pi/2} dx \, \log{(1+c \sin{x})} \\ &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} c^k \int_{-\pi/2}^{\pi/2} dx \, \sin^k{x} \\ &= - \frac{\pi}{2}\sum_{k=1}^{\infty} \frac{c^{2 k}}{k} \frac1{2^{2 k}} \binom{2 k}{k}   \end{align} $$
$$I'(c) = -\pi \sum_{k=1}^{\infty} \frac1{2^{2 k}} \binom{2 k}{k} c^{2 k-1} = -\frac{\pi}{c} \sum_{k=1}^{\infty} \frac1{2^{2 k}} \binom{2 k}{k} c^{2 k}= -\frac{\pi}{c} \left (\frac1{\sqrt{1-c^2}}-1\right )$$
$$\implies I(c) = -\pi \int dc \frac{1}{c} \left (\frac1{\sqrt{1-c^2}}-1\right ) = \pi \int dt \, (\sec{t}-\tan{t}) = \pi \log{(1+\sin{t})} + K$$
or
$$I(c) = K +\pi \log{\left ( 1+\sqrt{1-c^2} \right )} $$
$$I(0) = 0 \implies K=-\pi \log{2}$$
Therefore
$$I(c) = \pi \log{\left ( \frac{1+\sqrt{1-c^2}}{2}\right )} $$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
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\begin{align}&\color{#66f}{\large%
\int_{-\pi/2}^{\pi/2}\ln\pars{1 + c\sin\pars{x}}\,\dd x}
=\int_{-\pi/2}^{\pi/2}\ \overbrace{%
\int_{0}^{1}\frac{c\sin\pars{x}\,\dd t}{1 + c\sin\pars{x}t}}
^{\dsc{\ln\pars{1 + c\sin\pars{x}}}}\ \,\dd x
\\[5mm]&=\int_{0}^{1}
\int_{-\pi/2}^{\pi/2}
\frac{\bracks{1 + ct\sin\pars{x}} - 1\,\dd x}{1 + ct\sin\pars{x}}\,\frac{\dd t}{t}
=\int_{0}^{1}\bracks{\pi -\ \overbrace{\int_{-\pi/2}^{\pi/2}
\frac{\,\dd x}{1 + ct\sin\pars{x}}}
^{\ds{\dsc{\xi} = \dsc{\tan\pars{\frac{x}{2}}}}}}\,\frac{\dd t}{t}
\\[5mm]&=\int_{0}^{1}\bracks{\pi - \int_{1}^{1}
\frac{1}{1 + ct\pars{2\xi}/\pars{1 + \xi^{2}}}\,
\frac{2\,\dd\xi}{1 + \xi^{2}}}\,\frac{\dd t}{t}
=2\int_{0}^{1}
\bracks{\frac{\pi}{2} - \int_{-1}^{1}\frac{\dd\xi}{\xi^{2} + 2ct\xi + 1}}\,
\frac{\dd t}{t}
\\[5mm]&=2\int_{0}^{1}\bracks{%
{\pi \over 2}
-\int_{-1 + ct}^{1 + ct}\frac{\dd\xi}{\xi^{2} + 1 - c^{2}t^{2}}}\,\frac{\dd t}{t}
\\[5mm]&=2\int_{0}^{1}\braces{{\pi \over 2}
-\frac{1}{\root{1 - c^{2}t^{2}}}\ \overbrace{\bracks{%
\arctan\pars{\frac{1 + ct}{\root{1 - c^{2}t^{2}}}}
-\arctan\pars{\frac{-1 + ct}{\root{1 - c^{2}t^{2}}}}}}^{\ds{=\dsc{\frac{\pi}{2}}}}}
\,\frac{\dd t}{t}
\\[5mm]&=\pi\int_{0}^{1}\bracks{1
-\frac{1}{\root{1 - c^{2}t^{2}}}}\,\frac{\dd t}{t}
=\color{#66f}{\large\pi\ln\pars{\frac{1 + \root{1 - c^{2}}}{2}}}
\end{align}

Note that
\begin{align}&\overbrace{%
\int\frac{1}{\root{1 - c^{2}t^{2}}}\,\frac{\dd t}{t}}^{\ds{\dsc{ct}=\dsc{\cos\pars{\theta}}}}\ =\
\int\frac{1}{\sin\pars{\theta}}\,\frac{-\sin\pars{\theta}\,\dd\theta/c}{\cos\pars{\theta}/c}
=-\int\sec\pars{\theta}\,\dd\theta
\\[5mm]&=-\ln\pars{\sec\pars{\theta} + \tan\pars{\theta}}
=-\ln\pars{\frac{1 + \root{1 - \cos^{2}\pars{\theta}}}{\cos\pars{\theta}}}
\\[5mm]&=-\ln\pars{\frac{1 + \root{1 - c^{2}t^{2}}}{ct}} + \mbox{a constant}
\end{align}
such that
\begin{align}
&\pi\int_{0}^{1}\bracks{1
-\frac{1}{\root{1 - c^{2}t^{2}}}}\,\frac{\dd t}{t}
=\pi\bracks{\ln\pars{t} + \ln\pars{\frac{1 + \root{1 - c^{2}t^{2}}}{ct}}}_{0}^{1}
\\[5mm]&=\left.\pi\ln\pars{\frac{1 + \root{1 - c^{2}t^{2}}}{c}}
\right\vert_{\, 0}^{\, 1}=\pi\ln\pars{\frac{1 + \root{1 - c^{2}}}{2}}
\end{align}
A: An instant idea:
Introduce the function $$f(c) =\int_{-\pi/2}^{\pi/2}{\log(1+c\sin(x))dx}$$
Which satisfies the initial value $f(0) =0$
Next up is to differentiate under the integral sign and obtain a closed form of the derivative.
A: THIS IS ONLY A PARTIAL ANSWER
First we apply DUIS $$I'(a)=\int_{-\pi/2}^{\pi/2} \frac{\sin(x)}{a\sin(x)+1}$$
This can be integrated with a Weierstrass substitution, which I'll leave for you to verify:
$$\large\left.I'(a)=\frac{x-\frac{2\tan^{-1}\left(\frac{\tan(x/2)+y}{\sqrt{1-y^2}}\right) }{\sqrt{1-y^2}}}{y}\right|_{-\pi/2}^{\pi/2}$$
After plugging in the bounds we get:
$$\large I'(a)=\frac{\pi/2-\frac{2\tan^{-1}\left(\frac{1+y}{\sqrt{1-y^2}}\right)}{\sqrt{1-y^2}}}{y}+\frac{\pi/2+\frac{2\tan^{-1}\left(\frac{-1+y}{\sqrt{1-y^2}}\right)}{\sqrt{1-y^2}}}{y}=I_1+I_2$$
Now we have to integrate that huge monster.
$$\large I_1=\int\frac{\pi/2-\frac{2\tan^{-1}\left(\frac{1+y}{\sqrt{1-y^2}}\right)}{\sqrt{1-y^2}}}{y}$$
$$\large=\int\cos\theta\frac{\pi/2-\frac{2\tan^{-1}\left(\frac{1+\sin\theta}{\cos\theta}\right)}{\cos\theta}}{\sin\theta}(y=\sin\theta)$$
$$\large=\int\cot\theta\left({\pi/2-\frac{2\tan^{-1}\left(\frac{1+\sin\theta}{\cos\theta}\right)}{\cos\theta}}\right)$$
$$\large=\int\frac{\pi\cot\theta}2 +\int \frac{2\tan^{-1}\left(\frac{1+\sin\theta}{\cos\theta}\right)}{\sin\theta}$$
$$\large=\frac \pi 2 \log \sin\theta +\int \frac{2\tan^{-1}\left(\frac{1+\sin\theta}{\cos\theta}\right)}{\sin\theta}$$
Which is not expressible in terms of elementary functions.
A: Let $c=\sin a $\begin{align}
\int_{-\pi/2}^{\pi/2} \ln(1+\sin a\sin x)dx
=& \int_{-\pi/2}^{\pi/2}\int_0^a \frac{\cos t\sin x}{1+\sin t\sin x}dt\ dx \\
=& -\pi\int_0^a \tan\frac t2\ dt=2\pi\ln \left(\cos \frac a2\right)\\
=&\ \pi\ln\frac{1+\sqrt{1-c^2}}2
\end{align}
