Evaluate:$$H=\int^{\pi/2}_0 x\sin^3(x)\ln\left({\frac{1+\cos(x)}{\sin(x)}}\right)dx$$
Let: $$H(a)=\int^{\pi/2}_0 x\sin^3(x)\ln\left({\frac{1+a\cos(x)}{a\sin(x)}}\right)dx$$
Therfore: $$H'(a)=-\frac{1}{a}\int^{\pi/2}_0 \frac{x\sin^3(x)}{1+a\cos(x)}dx\overset{y=\cos(x)}{=}-\frac{1}{a}\int^{1}_0 \frac{\arccos(y)(1-y^2)}{1+ay}dy$$ Need help evaluating this integration?
Let $$I = \int_{0}^{\pi/2} x\sin^3(x)\ln\left({\frac{1+\cos(x)}{\sin(x)}}\right)dx$$
Therefore,
\begin{align} I &= \int_{0}^{\pi/2} x \left(\frac{3\sin(x) - \sin(3x)}{4}\right)\ln(\cot(x/2))dx \\ &= \frac{3}{4}\int_{0}^{\pi/2} x\sin(x)\ln(\cot(x/2))dx - \frac{1}{4}\int_{0}^{\pi/2} x\sin(3x)\ln(\cot(x/2))dx \\ &= \frac{3}{4}I_1 - \frac{1}{4}I_3 \end{align}
Solving for $I_1$. Let $u = \ln(\cot(2/x)) \implies du = -\csc(x)dx$ and let $dv = x\sin(x)dx \implies v = \sin(x) - x\cos(x)$. Therefore,
\begin{align} I_1 &= \int_{0}^{\pi/2} x\sin(x)\ln(\cot(x/2)) \, dx \\ &= \left[(\sin(x)-x\cos(x))\ln(\cot(x/2))\right]_0^{\pi/2} - \int_0^{\pi/2} (\sin(x)-x\cos(x))(-\csc(x)) \, dx \\ &= 0 + \int_0^{\pi/2} (1-x\cot(x))\, dx \\ &= \frac{\pi}{2} - \int_0^{\pi/2}x\cot(x) \, dx \\ &= \frac{\pi}{2} - \frac{\pi}{2}\ln 2 \end{align}
Here, $\int_0^{\pi/2}x\cot(x) \, dx = \frac{\pi}{2}\ln 2$
Solving for $I_3$. Let $u = \ln(\cot(2/x)) \implies du = -\csc(x)dx$. Let $dv = x\sin(3x)dx \implies v = \frac{\sin(3x)}{9}-\frac{x\cos(3x)}{3}$. Therefore,
\begin{align} I_3 &= \int_{0}^{\pi/2} x\sin(3x)\ln(\cot(x/2))dx \\ &= \left[\left(\frac{\sin(3x)}{9}-\frac{x\cos(3x)}{3}\right)\ln(\cot(x/2))\right]_0^{\pi/2} + \int_0^{\pi/2} \left(\frac{\sin(3x)}{9\sin(x)} - \frac{x\cos(3x)}{3\sin(x)}\right)dx \\ &= \int_0^{\pi/2} \frac{3-4\sin^2(x)}{9}dx - \frac{1}{3}\int_0^{\pi/2}x\left(\frac{\cos(3x)}{\sin(x)}\right)dx \\ &= \int_0^{\pi/2} \left(\frac{1}{3} - \frac{4}{9}\sin^2x\right)dx - \frac{1}{3}\int_0^{\pi/2}x\cot x(1-4\sin^2x)dx \\ &= \left[\frac{x}{3} - \frac{4}{9}\left(\frac{x}{2}-\frac{\sin(2x)}{4}\right)\right]_0^{\pi/2} - \frac{1}{3}\left[\int_0^{\pi/2} x\cot x \, dx - 2\int_0^{\pi/2} x\sin(2x)dx \right] \\ &= \left[\frac{\pi}{6} - \frac{2}{9}\left(\frac{\pi}{2}\right)\right] - \frac{1}{3}\left[ \frac{\pi}{2}\ln 2 - 2\left(\frac{\pi}{4}\right)\right] \\ &= \frac{2\pi}{9} - \frac{\pi}{6}\ln 2 \end{align}
Combining $I_1$ and $I_3$ we get:
\begin{align} I &= \frac{3}{4}I_1 - \frac{1}{4}I_3 \\ &= \frac{3}{4}\left(\frac{\pi}{2} - \frac{\pi}{2}\ln 2\right) - \frac{1}{4}\left(\frac{2\pi}{9} - \frac{\pi}{6}\ln 2\right) \\ &= \frac{\pi}{72}(23 - 24\ln 2) \end{align}
Which matches the answer given by Wolfram.
We have to evaluate
$$ H = \int_{0}^{\pi/2}x\sin^3x\;\ln\!\Bigl(\tfrac{1+\cos x}{\sin x}\Bigr)\,dx. $$
First observe the neat identity
$$ \frac{1+\cos x}{\sin x} =\frac{2\cos^2\frac x2}{2\sin\frac x2\cos\frac x2} =\cot\!\Bigl(\tfrac x2\Bigr). $$
Hence
$$ H=\int_0^{\pi/2}x\sin^3x\;\ln\!\bigl(\cot\tfrac x2\bigr)\,dx. $$
Let
$$ u = \ln\!\bigl(\cot\tfrac x2\bigr), \quad dv = x\sin^3x\,dx. $$
We compute
$$ du = \frac{d}{dx}\ln\!\bigl(\cot\tfrac x2\bigr)\,dx = -\frac{dx}{\sin x}. $$
(One shows $\tfrac d{dx}\ln(\cot(x/2))=-1/\sin x$ by the half‑angle formulas.)
$$ \int x\sin^3x\,dx = -x\cos x+\frac{x\cos^3x}{3}+\frac{2}{3}\sin x+\frac{1}{9}\sin^3x + C. $$
Denote this by $v(x)$.
Integration by parts on $[0,\pi/2]$ gives
$$ H=\Bigl[u\,v\Bigr]_0^{\pi/2} - \int_0^{\pi/2}v\,du. $$
One checks $u\,v\to0$ at both endpoints, so the boundary term vanishes and
$$ H=\int_0^{\pi/2}\!v(x)\,\frac{dx}{\sin x}. $$
Writing out $v(x)$ and dividing by $\sin x$ yields
$$ H =\int_0^{\pi/2}\Bigl[-x\cot x+\tfrac{x\cos^3x}{3\sin x}+\tfrac23+\tfrac19\sin^2x\Bigr]\,dx. $$
Split this as
$$ H =\underbrace{\int_0^{\pi/2}\Bigl[-x\cot x+\tfrac{x\cos^3x}{3\sin x}\Bigr]dx}_{(*)} \;+\;\int_0^{\pi/2}\Bigl(\tfrac23+\tfrac19\sin^2x\Bigr)dx. $$
Hence
$$ \int_0^{\pi/2}\Bigl(\tfrac23+\tfrac19\sin^2x\Bigr)dx =\frac\pi3+\frac\pi{36}=\frac{13\pi}{36}. $$
Combine the two terms under one denominator:
$$ -x\cot x+\frac{x\cos^3x}{3\sin x} =x\,\frac{\cos^3x-3\cos x}{3\sin x} =-x\Bigl(\tfrac13\sin x\cos x+\tfrac23\cot x\Bigr). $$
Thus
$$ (*)=-\frac13\int_0^{\pi/2}x\sin x\cos x\,dx -\frac23\int_0^{\pi/2}x\cot x\,dx. $$
We evaluate each in turn:
Use $\sin x\cos x=\tfrac12\sin2x$ and integrate by parts:
$$ I_1 = \frac12\int_0^{\pi/2}x\sin2x\,dx = \frac12\cdot\frac\pi4 = \frac\pi8. $$
Integrate by parts with $\cot x=(\ln\sin x)'$. One finds
$$ I_2 = x\ln(\sin x)\Big|_0^{\pi/2} -\int_0^{\pi/2}\ln(\sin x)\,dx = -\!\int_0^{\pi/2}\ln(\sin x)\,dx = \frac\pi2\ln2. $$
(The last line is the classic $\int_0^{\pi/2}\ln\sin x\,dx=-\tfrac\pi2\ln2$.)
Putting them together,
$$ (*)=-\frac13\cdot\frac\pi8\;-\;\frac23\cdot\frac\pi2\ln2 =\,-\frac\pi{24}\;-\;\frac\pi3\ln2. $$
Adding the two pieces,
$$ H =\Bigl[\tfrac{13\pi}{36}\Bigr] \;+\;\Bigl[-\tfrac\pi{24}-\tfrac\pi3\ln2\Bigr] =\frac{23\pi}{72}-\frac\pi3\ln2. $$
Hence the integral has the closed form
$$ \boxed{ \int_{0}^{\pi/2} x\sin^3x\;\ln\!\Bigl(\tfrac{1+\cos x}{\sin x}\Bigr)\,dx =\frac{23\pi}{72}\;-\;\frac{\pi}{3}\ln2. } $$
By @polychroma ' s comment, we have $$ \ln(\csc(x)+\cot(x))=\sum_{n=0}^{\infty}\frac{2\cos((2n+1)x)}{2n+1} $$ And so we have $$ H = \sum_{n=0}^{\infty}\frac{2}{2n+1}\int_0^{\pi/2}\cos((2n+1)x)x\sin^3(x)dx \quad (1) $$ assuming uniform convergence.
Then we want to compute the definite integral $$I_n = \int_{0}^{\pi/2} x \sin^3(x) \cos((2n+1)x) \, dx.$$
First we use the triple angle formula for sine: $\sin(3x) = 3\sin(x) - 4\sin^3(x)$.
Substitute this into the integral: $$ I_n = \int_{0}^{\pi/2} x \left( \frac{3\sin(x) - \sin(3x)}{4} \right) \cos((2n+1)x) \, dx \\ = \frac{1}{4} \int_{0}^{\pi/2} x \left[ 3\sin(x)\cos((2n+1)x) - \sin(3x)\cos((2n+1)x) \right] \, dx $$
Now recall the product-to-sum identity: $$\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)].$$
For the first term: $3\sin(x)\cos((2n+1)x) = \frac{3}{2}[\sin((2n+2)x) + \sin(-2nx)] = \frac{3}{2}[\sin((2n+2)x) - \sin(2nx)]$
For the second term: $-\sin(3x)\cos((2n+1)x) = -\frac{1}{2}[\sin((2n+4)x) + \sin((2-2n)x)]$
Substitute back: $$ I_n = \frac{1}{4} \int_{0}^{\pi/2} x \left[ \frac{3}{2}\sin((2n+2)x) - \frac{3}{2}\sin(2nx) - \frac{1}{2}\sin((2n+4)x) - \frac{1}{2}\sin(2(1-n)x) \right] \, dx =\\ \frac{1}{8} \int_{0}^{\pi/2} x \left[ 3\sin((2n+2)x) - 3\sin(2nx) - \sin((2n+4)x) - \sin(2(1-n)x) \right] \, dx $$
Let $J_k = \int_{0}^{\pi/2} x \sin(kx) dx$. IBP implies $$J_k = \left[ -\frac{x}{k}\cos(kx) \right]_{0}^{\pi/2} - \int_{0}^{\pi/2} -\frac{1}{k}\cos(kx) dx =\\ \left[ -\frac{x}{k}\cos(kx) + \frac{1}{k^2}\sin(kx) \right]_{0}^{\pi/2} =\\ -\frac{\pi}{2k}\cos(k\pi/2) + \frac{1}{k^2}\sin(k\pi/2)$$
Note that for $k = 2n+2, 2n, 2n+4, 2(1-n)$, when multiplied by $\pi/2$, they become $(n+1)\pi, n\pi, (n+2)\pi, (1-n)\pi$ respectively. Since $n$ is an integer, these are always integer multiples of $\pi$. Therefore, $\sin(k\pi/2) = 0$ for all these terms.
So, for $k \neq 0$: $J_k = -\frac{\pi}{2k}\cos(k\pi/2)$. Also, $\cos(m\pi) = (-1)^m$.
So for each $k$ we have
$3 J_{2n+2} = \frac{3\pi(-1)^n}{4(n+1)}$
$-3 J_{2n} = \frac{3\pi}{4n}(-1)^n$
$- J_{2n+4} = \frac{\pi(-1)^n}{4(n+2)}$
$- J_{2-2n} = \frac{\pi(-1)^n}{4(n-1)}$
Combining these results: $$ I_n = \frac{1}{8} \left[ \frac{3\pi(-1)^n}{4(n+1)} + \frac{3\pi(-1)^n}{4n} + \frac{\pi(-1)^n}{4(n+2)} + \frac{\pi(-1)^n}{4(n-1)} \right] =\\ \frac{\pi(-1)^n}{32} \left[ \frac{3}{n+1} + \frac{3}{n} + \frac{1}{n+2} + \frac{1}{n-1} \right] \quad \text{for } n \neq 0, 1 $$
For the special cases $n=0, 1$ we have
$n=0:$ $$I_0 = \frac{1}{8} \int_{0}^{\pi/2} x \left[ 3\sin(2x) - 3\sin(0x) - \sin(4x) - \sin(2(1-0)x) \right] \, dx=\\ \frac{1}{8} \int_{0}^{\pi/2} x \left[ 2\sin(2x) - \sin(4x) \right] \, dx = \frac{5\pi}{64}$$
$n=1:$ $$I_1 = \frac{1}{8} \int_{0}^{\pi/2} x \left[ 3\sin(4x) - 3\sin(2x) - \sin(6x) - \sin(0x) \right] \, dx =-\frac{29\pi}{192}$$
Now going back to (1) we still need to calculate $$ H = \sum_{n=0}^{\infty}\frac{2}{2n+1} I_n = 2I_0-2/3I_1+S \quad (2) $$ with $S := \sum_{n=2}^{\infty}\frac{2}{2n+1} I_n$.
The summands for $n\geq 2$ are given by $$ a_n = \frac{2}{(2n+1)} I_n= \frac{\pi(-1)^n (2n^2+2n-3)}{8(n-1)n(n+1)(n+2)} $$
After partial fraction decomposition this is given by
$$ a_n = \frac{\pi(-1)^n}{8} \left[ \frac{1}{6(n-1)} + \frac{3}{2n} - \frac{3}{2(n+1)} - \frac{1}{6(n+2)} \right] $$
We can split the sum into four parts: $$ S = \frac{\pi}{8} \left[ \frac{1}{6} \sum_{n=2}^{\infty} \frac{(-1)^n}{n-1} + \frac{3}{2} \sum_{n=2}^{\infty} \frac{(-1)^n}{n} - \frac{3}{2} \sum_{n=2}^{\infty} \frac{(-1)^n}{n+1} - \frac{1}{6} \sum_{n=2}^{\infty} \frac{(-1)^n}{n+2} \right] $$ Each of these series can be computed using the known sum of the alternating harmonic series: $$\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \ln(2).$$
In particular:
$\sum_{n=2}^{\infty} \frac{(-1)^n}{n-1}= \ln(2)$.
$\sum_{n=2}^{\infty} \frac{(-1)^n}{n}= 1 - \ln(2)$.
$\sum_{n=2}^{\infty} \frac{(-1)^n}{n+1}= -\ln(2) + \frac{1}{2}$. So, $\sum_{n=2}^{\infty} \frac{(-1)^n}{n+1} = -(-\ln(2) + \frac{1}{2}) = \ln(2) - \frac{1}{2}$.
$\sum_{n=2}^{\infty} \frac{(-1)^n}{n+2}=-\ln(2) + \frac{5}{6}$.
$$ S = \frac{\pi}{8} \left[ \frac{-24\ln(2) + 19}{9} \right] = \frac{\pi(19 - 24\ln(2))}{72} $$
Now inserting this back in (2) and having $I_0 = 5 \frac{\pi}{64}, I_1 = -29 \frac{\pi}{192}$
we get the final answer
$$
\boxed{H = \frac{\pi }{18}+\frac{1}{72} \pi (19-24 \log (2)).}
$$
