Question Number 165380 by mnjuly1970 last updated on 31/Jan/22
$$ \\ $$$$\:\:\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\:{x}^{\:\mathrm{3}} }{{sin}^{\:\mathrm{2}} \left({x}\right)}{dx}\overset{?} {=}\:\frac{\mathrm{3}}{\mathrm{8}}\:\left(\pi^{\:\mathrm{2}} {ln}\left(\mathrm{4}\right)−\mathrm{7}\zeta\left(\mathrm{3}\right)\right) \\ $$
Answered by Ar Brandon last updated on 31/Jan/22
![I=∫_0 ^(π/2) x^3 cosec^2 xdx=−[x^3 cotx]_0 ^(π/2) +3∫_0 ^(π/2) x^2 cotxdx =3[x^2 ln(sinx)]_0 ^(π/2) −6∫_0 ^(π/2) xln(sin(x)dx](https://www.tinkutara.com/question/Q165384.png)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}^{\mathrm{3}} \mathrm{cosec}^{\mathrm{2}} {xdx}=−\left[{x}^{\mathrm{3}} \mathrm{cot}{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\mathrm{3}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}^{\mathrm{2}} \mathrm{cot}{xdx} \\ $$$$\:\:\:=\mathrm{3}\left[{x}^{\mathrm{2}} \mathrm{ln}\left(\mathrm{sin}{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\mathrm{6}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\mathrm{ln}\left(\mathrm{sin}\left({x}\right){dx}\right. \\ $$