Question Number 114107 by mathdave last updated on 17/Sep/20
![prove that ∫_0 ^(π/2) [((ln(((1−sinx)/(1+sinx)))(√(cosx)))/((1+sinx)(√(1−sinx))))]dx=−8](https://www.tinkutara.com/question/Q114107.png)
$${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[\frac{\mathrm{ln}\left(\frac{\mathrm{1}−\mathrm{sin}{x}}{\mathrm{1}+\mathrm{sin}{x}}\right)\sqrt{\mathrm{cos}{x}}}{\left(\mathrm{1}+\mathrm{sin}{x}\right)\sqrt{\mathrm{1}−\mathrm{sin}{x}}}\right]{dx}=−\mathrm{8} \\ $$
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
![=∫_0 ^(π/2) ((ln(((1−cos(x))/(1+cos(x))))(√(sin(x))))/((1+cos(x))(√(1−cos(x)))))dx 1+cos(x)=2cos^2 ((x/2)) and 1−cos(x)=2sin^2 ((x/2))⇒ =∫_0 ^(π/2) ((ln(tg^2 ((x/2)))(√(2sin((x/2))cos((x/2)))))/(2cos^2 ((x/2))(√(2sin^2 ((x/2))))))dx =2∫_0 ^(π/2) ((ln(tg((x/2))))/( (√((sin((x/2)))/(cos((x/2)))))))(dx/(2cos^2 ((x/2)))) tg((x/2))=t⇒dt=(dx/(2cos^2 ((x/2)))) we get =2∫_0 ^1 ((ln(x))/( (√x)))dx=[4(√x)ln(x)]_0 ^1 −4∫_0 ^1 (dx/( (√x)))=−4[2(√x)]_0 ^1 =−8](https://www.tinkutara.com/question/Q114220.png)
$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left(\frac{\mathrm{1}−{cos}\left({x}\right)}{\mathrm{1}+{cos}\left({x}\right)}\right)\sqrt{{sin}\left({x}\right)}}{\left(\mathrm{1}+{cos}\left({x}\right)\right)\sqrt{\mathrm{1}−{cos}\left({x}\right)}}{dx} \\ $$$$\mathrm{1}+{cos}\left({x}\right)=\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\:{and}\:\mathrm{1}−{cos}\left({x}\right)=\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\Rightarrow \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left({tg}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\sqrt{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)}}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\sqrt{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left({tg}\left(\frac{{x}}{\mathrm{2}}\right)\right)}{\:\sqrt{\frac{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{{cos}\left(\frac{{x}}{\mathrm{2}}\right)}}}\frac{{dx}}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$${tg}\left(\frac{{x}}{\mathrm{2}}\right)={t}\Rightarrow{dt}=\frac{{dx}}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}\:{we}\:{get} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)}{\:\sqrt{{x}}}{dx}=\left[\mathrm{4}\sqrt{{x}}{ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\:\sqrt{{x}}}=−\mathrm{4}\left[\mathrm{2}\sqrt{{x}}\right]_{\mathrm{0}} ^{\mathrm{1}} =−\mathrm{8} \\ $$$$ \\ $$