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Question Number 137328 by mnjuly1970 last updated on 01/Apr/21

Answered by Dwaipayan Shikari last updated on 01/Apr/21

$$-{\int }_0^{\frac{\pi }{2}}{tan}^2\left(x\right)log\left(sinx\right)dx$$$$={\int }_0^{\frac{\pi }{2}}log\left(sinx\right)-{\int }_0^{\frac{\pi }{2}}{sec}^2\left(x\right)log\left(sinx\right)dx$$$$=-\frac{\pi }{2}log\left(2\right)-{\left[log\left(sinx\right)tan\left(x\right)\right]}_0^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}tan\left(x\right)cot\left(x\right)dx$$$$=\frac{-\pi }{2}log\left(2\right)-0+\frac{\pi }{2}=\frac{\pi }{2}log\left(\frac{e}{2}\right)$$

Commented by mnjuly1970 last updated on 01/Apr/21

$$excellent...thanksalot....$$

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