Two definite integrals in terms of one another

Question

While practicing definite integration I stumbled upon following question:

Q.) If $\int_{0}^{π/2}\ln(\sin x)dx=k$ then, find $\int_{0}^{π/2}\frac{x²}{(\sin x)²}dx$, in terms of $k$

My attempt: I have memorised the first integral answer as

$\int_{0}^{π/2}\ln(\sin x)dx$=$-\frac{π}{2}\ln 2$=$k$

For second integral I did following:

Let $$I=\int_{0}^{π/2}\frac{x²}{(\sin x)²}dx$$

$$I=\int_{0}^{π/2}\frac{(\frac{π}{2}-x)²}{(\cos x)²}dx$$

$$[\int_{b}^{a}f(x)dx=\int_{b}^{a}f(a+b-x)dx]$$

and then i did by parts, but things become only more complicated. So please help if you can solve the second integral easily. THANKS IN ADVANCE

'ANSWER IS $-2k$'

Answer 1

Hint: Start from your $k$ integral and integrate by parts twice.

Answer 2

Integrate by parts so that $$ \begin{aligned}\int_0^{\frac{\pi}{2}} \dfrac{x^2}{\sin^2 x} \mathrm{d}x &=\underbrace{-x^2 \cot x \bigg|_0^{\frac{\pi}{2}}}_0+2\int_0^{\frac{\pi}{2}}x\cot x \mathrm{d}x \\&=2\left(\underbrace{x \ln \sin x \bigg|_0^{\frac{\pi}{2}}}_0-\int_0^{\frac{\pi}{2}}\ln \sin x \mathrm{d}x\right) \\ &=-2k \\ &=\pi\ln 2 \end{aligned}$$