Evaluating $\int_0^{\pi/2}({x \over \sin x})^2dx$ using value of a given integral

Question

Question

If $\int_0^{\pi/2}\ln({\sin x})dx = {\pi \over 2}\ln({1\over2})$ then find the value of $\int_0^{\pi/2}({x \over \sin x})^2dx$

I'm stumped. I have no clue what to do. A hint would be appreciated.

Answer 1

• First question, before the edit.

  One may observe that $$ \int \frac1{\sin x}\:dx=\ln (\tan(x/2))+C. $$ Then one may use an integration by parts, $$ \begin{align} \int_0^{\pi/2}\frac{x}{\sin x}\:dx&=\left[x\ln (\tan(x/2))\right]_0^{\pi/2}-\int_0^{\pi/2} \ln (\tan(x/2))dx \\\\&=0-2\int_0^{\pi/4} \ln (\tan(u))du \\\\&=-2\int_0^1 \frac{\ln v\:dv}{1+v^2}\quad (v=\tan u) \\\\&=2\sum_{n=0}^\infty(-1)^{n+1}\int_0^1 v^{2n}\ln v \:dv \\\\&=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2} \\\\&=2C \end{align} $$ where $C$ is the Catalan constant.

• Second question, after the edit.

  One may observe that $$ \int \frac{-1}{(\sin x)^2}\:dx=\int \frac{(\cos x)'(\sin x)-(\sin x)'(\cos x)}{(\sin x)^2}\:dx=\int \left(\frac{\cos x}{\sin x}\right)'dx=\frac{\cos x}{\sin x}+C. $$ Then one may integrate by parts twice, $$ \begin{align} \int_0^{\pi/2}\left(\frac{x}{\sin x}\right)^2dx&=\left[-x^2\frac{\cos x}{\sin x}\right]_0^{\pi/2}+2\int_0^{\pi/2} x\:\frac{\cos x}{\sin x}dx \\\\&=0+2\left(\left[x\ln(\sin x)\right]_0^{\pi/2}-\int_0^{\pi/2} \ln(\sin x)\:dx\right) \\\\&=\pi\ln 2 \end{align} $$ using the assumed result.

Answer 2

HINT: Please fix the question. It should be $\int_0^{\pi/2}(\frac{x}{\sin x})^2\ dx$ and the answer is $\pi\ln2$

FURTHER HINT: If we differentiate $x^2\cot x$ we get $2x\cot x-\frac{x^2}{\sin^2x}$. If we differentiate $x\ln\sin x$ we get $x\cot x+\ln\sin x$. And we are given the integral (over the relevant range) of $\ln\sin x$).

I agree that the tricky part is then evaluating $2x\ln\sin x$ and $x^2\cot x$ at $x=0$. You need to take the limit as $x\to 0$. They are 0 in both cases.