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Question Number 154458 by mathdanisur last updated on 18/Sep/21

Answered by ARUNG_Brandon_MBU last updated on 18/Sep/21

$$\int \frac{x^2}{{\sin }^{2}x}dx=-x^2\cot x+2\int x\cot xdx$$$$=-x^2\cot x+2x\ln \left(\sin x\right)-2\int \ln \left(\sin x\right)dx$$$${\int }_0^{\frac{\pi }{2}}\ln \left(\sin x\right)dx=-\frac{\pi \ln 2}{2}$$$${\int }_0^{\frac{\pi }{4}}\ln \left(\sin x\right)dx=-\frac{G}{2}-\frac{\pi \ln 2}{4}$$$${\int }_0^{\frac{\pi }{6}}\ln \left(\sin x\right)dx=$$

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