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Question Number 114107 by mathdave last updated on 17/Sep/20

$$provethat$$$${\int }_0^{\frac{\pi }{2}}\left[\frac{\ln \left(\frac{1-\sin x}{1+\sin x}\right)\sqrt{\cos x}}{(1+\sin x)\sqrt{1-\sin x}}\right]dx=-8$$

Commented by mathdave last updated on 17/Sep/20

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

Answered by maths mind last updated on 17/Sep/20

$$={\int }_0^{\frac{\pi }{2}}\frac{ln\left(\frac{1-cos\left(x\right)}{1+cos\left(x\right)}\right)\sqrt{sin\left(x\right)}}{(1+cos\left(x\right))\sqrt{1-cos\left(x\right)}}dx$$$$1+cos\left(x\right)=2{cos}^2\left(\frac{x}{2}\right)and1-cos\left(x\right)=2{sin}^2\left(\frac{x}{2}\right)\Rightarrow$$$$={\int }_0^{\frac{\pi }{2}}\frac{ln\left({tg}^2\left(\frac{x}{2}\right)\right)\sqrt{2sin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\right)}}{2{cos}^2\left(\frac{x}{2}\right)\sqrt{2{sin}^2\left(\frac{x}{2}\right)}}dx$$$$=2{\int }_0^{\frac{\pi }{2}}\frac{ln\left(tg\left(\frac{x}{2}\right)\right)}{\sqrt{\frac{sin\left(\frac{x}{2}\right)}{cos\left(\frac{x}{2}\right)}}}\frac{dx}{2{cos}^2\left(\frac{x}{2}\right)}$$$$tg\left(\frac{x}{2}\right)=t\Rightarrow dt=\frac{dx}{2{cos}^2\left(\frac{x}{2}\right)}weget$$$$=2{\int }_0^1\frac{ln\left(x\right)}{\sqrt{x}}dx={\left[4\sqrt{x}ln\left(x\right)\right]}_0^1-4{\int }_0^1\frac{dx}{\sqrt{x}}=-4{\left[2\sqrt{x}\right]}_0^1=-8$$$$$$

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