Question Number 137328 by mnjuly1970 last updated on 01/Apr/21
Answered by Dwaipayan Shikari last updated on 01/Apr/21
![−∫_0 ^(π/2) tan^2 (x)log(sinx)dx =∫_0 ^(π/2) log(sinx)−∫_0 ^(π/2) sec^2 (x)log(sinx)dx =−(π/2)log(2)−[log(sinx)tan(x)]_0 ^(π/2) +∫_0 ^(π/2) tan(x)cot(x)dx =((−π)/2)log(2)−0+(π/2)=(π/2)log((e/2))](https://www.tinkutara.com/question/Q137330.png)
$$−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {tan}^{\mathrm{2}} \left({x}\right){log}\left({sinx}\right){dx}\:\:\:\:\:\: \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sinx}\right)−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sec}^{\mathrm{2}} \left({x}\right){log}\left({sinx}\right){dx} \\ $$$$=−\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right)−\left[{log}\left({sinx}\right){tan}\left({x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {tan}\left({x}\right){cot}\left({x}\right){dx} \\ $$$$=\frac{−\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right)−\mathrm{0}+\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}{log}\left(\frac{{e}}{\mathrm{2}}\right) \\ $$
Commented by mnjuly1970 last updated on 01/Apr/21
$$\:{excellent}…{thanks}\:{alot}…. \\ $$