Question Number 66090 by Lalita 42@gmeilcon last updated on 09/Aug/19
$$\:\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:{dx}\:= \\ $$
Answered by mr W last updated on 09/Aug/19
![I=∫_( 0) ^(π/2) ((sin^2 x)/(sin x+cos x)) dx let x=(π/2)−t I=−∫_( (π/2)) ^0 ((cos^2 t)/(cos t+sin t)) dt I=∫^(π/2) _ _0 ((cos^2 t)/(cos t+sin t)) dt I=∫^(π/2) _ _0 ((1−sin^2 t)/(cos t+sin t)) dt I=∫^(π/2) _ _0 (1/(cos t+sin t)) dt−I 2I=∫^(π/2) _ _0 (1/(cos t+sin t)) dt =(1/( (√2)))[ln ∣((tan (t/2)+(√2)−1)/(tan (t/2)−(√2)−1))∣]_0 ^(π/2) =(1/( (√2)))[ln ∣((1+(√2)−1)/(1−(√2)−1))∣−ln ∣(((√2)−1)/(−(√2)−1))∣] =(2/( (√2)))ln ((√2)+1) ⇒I=(1/( (√2)))ln ((√2)+1)](https://www.tinkutara.com/question/Q66094.png)
$$\:{I}=\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:{dx} \\ $$$${let}\:{x}=\frac{\pi}{\mathrm{2}}−{t} \\ $$$$\:{I}=−\underset{\:\frac{\pi}{\mathrm{2}}} {\overset{\mathrm{0}} {\int}}\:\frac{\mathrm{cos}^{\mathrm{2}} \:{t}}{\mathrm{cos}\:{t}+\mathrm{sin}\:{t}}\:{dt} \\ $$$$\:{I}=\underset{\:_{\mathrm{0}} } {\int}^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{cos}^{\mathrm{2}} \:{t}}{\mathrm{cos}\:{t}+\mathrm{sin}\:{t}}\:{dt} \\ $$$$\:{I}=\underset{\:_{\mathrm{0}} } {\int}^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{t}}{\mathrm{cos}\:{t}+\mathrm{sin}\:{t}}\:{dt} \\ $$$$\:{I}=\underset{\:_{\mathrm{0}} } {\int}^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}}{\mathrm{cos}\:{t}+\mathrm{sin}\:{t}}\:{dt}−{I} \\ $$$$\:\mathrm{2}{I}=\underset{\:_{\mathrm{0}} } {\int}^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}}{\mathrm{cos}\:{t}+\mathrm{sin}\:{t}}\:{dt} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\mathrm{ln}\:\mid\frac{\mathrm{tan}\:\frac{{t}}{\mathrm{2}}+\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{tan}\:\frac{{t}}{\mathrm{2}}−\sqrt{\mathrm{2}}−\mathrm{1}}\mid\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\mathrm{ln}\:\mid\frac{\mathrm{1}+\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{2}}−\mathrm{1}}\mid−\mathrm{ln}\:\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{−\sqrt{\mathrm{2}}−\mathrm{1}}\mid\right] \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}\mathrm{ln}\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{ln}\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$