Question Number 223920 by Mathspace last updated on 09/Aug/25
$${find}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}^{\mathrm{2}} }{{sin}^{\mathrm{2}} {x}}{dx} \\ $$
Answered by Frix last updated on 09/Aug/25
![by parts (2 times) ∫_0 ^(π/2) (x^2 /(sin^2 x))dx= =[−(x^2 /(tan x))+2xln sin x]_0 ^(π/2) −2∫_0 ^(π/2) ln sin x dx= =0+πln 2= =πln 2](https://www.tinkutara.com/question/Q223922.png)
$$\mathrm{by}\:\mathrm{parts}\:\left(\mathrm{2}\:\mathrm{times}\right) \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{x}^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$=\left[−\frac{{x}^{\mathrm{2}} }{\mathrm{tan}\:{x}}+\mathrm{2}{x}\mathrm{ln}\:\mathrm{sin}\:{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{ln}\:\mathrm{sin}\:{x}\:{dx}= \\ $$$$=\mathrm{0}+\pi\mathrm{ln}\:\mathrm{2}= \\ $$$$=\pi\mathrm{ln}\:\mathrm{2} \\ $$