Question Number 177298 by peter frank last updated on 03/Oct/22
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)}\mathrm{dx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{log}\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$
Answered by Ar Brandon last updated on 11/Oct/22
![I=∫_0 ^(π/2) ((sin^2 x)/(sinx+cosx))dx ...eqn(i) I=∫_0 ^(π/2) ((cos^2 x)/(sinx+cosx))dx ...eqn(ii) [letting t=(π/2)−x] eqn(i)+eqn(ii) ⇒ 2I=∫_0 ^(π/2) ((sin^2 x+cos^2 x)/(sinx+cosx))dx=∫_0 ^(π/2) (1/(sinx+cosx))dx I=(1/2)∫_0 ^(π/2) (1/(sinx+cosx))dx=(1/2)∫_0 ^1 (1/(((2t)/(1+t^2 ))+((1−t^2 )/(1+t^2 ))))∙(2/(1+t^2 ))dt =∫_0 ^1 (1/(1+2t−t^2 ))dt=∫_0 ^1 (dt/(2−(t−1)^2 ))=(1/( (√2)))[argth(((t−1)/( (√2))))]_0 ^1 =(1/(2(√2)))[ln∣((t+(√2)−1)/(t−(√2)−1))∣]_0 ^1 =(1/(2(√2)))ln∣(((√2)+1)/( (√2)−1))∣=(1/(2(√2)))ln((√2)+1)^2 =(1/( (√2)))ln((√2)+1)](https://www.tinkutara.com/question/Q177299.png)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{sin}{x}+\mathrm{cos}{x}}{dx}\:…\mathrm{eqn}\left({i}\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}^{\mathrm{2}} {x}}{\mathrm{sin}{x}+\mathrm{cos}{x}}{dx}\:\:\:…\mathrm{eqn}\left({ii}\right)\:\left[\mathrm{letting}\:{t}=\frac{\pi}{\mathrm{2}}−{x}\right] \\ $$$$\mathrm{eqn}\left({i}\right)+\mathrm{eqn}\left({ii}\right)\:\Rightarrow \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{cos}^{\mathrm{2}} {x}}{\mathrm{sin}{x}+\mathrm{cos}{x}}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{sin}{x}+\mathrm{cos}{x}}{dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{sin}{x}+\mathrm{cos}{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\centerdot\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} }{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{2}−\left({t}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\mathrm{argth}\left(\frac{{t}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left[\mathrm{ln}\mid\frac{{t}+\sqrt{\mathrm{2}}−\mathrm{1}}{{t}−\sqrt{\mathrm{2}}−\mathrm{1}}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\mid\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\mid=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$
Commented by peter frank last updated on 03/Oct/22
$$\mathrm{thank}\:\mathrm{you} \\ $$