Question

${\int }_0^{\pi /2}\frac{{\sin }^{2}x}{\sin x+\cos x}dx=$

• A. $\sqrt{2}\log (\sqrt{2}+1)$
• B. $\frac{1}{\sqrt{2}}log(\sqrt{2}+1)$
• C. $\log (\sqrt{2}+1)$
• D. $\frac{1}{\sqrt{2}}\log (\sqrt{2}-1)$

Answer 1

The correct option is B $\frac{1}{\sqrt{2}}log(\sqrt{2}+1)$

$I={\int }_0^{\frac{\pi }{2}}\frac{si{n}^{2}x}{sinx+cosx}dx$
$I={\int }_0^{\frac{\pi }{2}}\frac{co{s}^2}{sinx+cosx}dx$
$2I={\int }_0^{\frac{\pi }{2}}\frac{1}{sinx+cosx}dx$
$2I={\int }_0^{\frac{\pi }{2}}\frac{(1+ta{n}^2\frac{x}{2})}{-ta{n}^2\frac{x}{2}+2tan\frac{x}{2}+1}dx$
$2I={\int }_0^{\frac{\pi }{2}}\frac{2dtan\frac{x}{2}}{(\sqrt{2}{)}^2{(tan\frac{x}{2}-1)}^2}$
$=\frac{1}{2\sqrt{2}}log\left|\frac{\sqrt{2}+ta{n}^n\left(\frac{x}{2}\right)-1}{\sqrt{2}\left(tan\frac{x}{2}\right)+1}\right|$
$=\frac{1}{2\sqrt{2}}log\left|\frac{ta{n}^n\left(\frac{x}{2}\right)+\sqrt{2}-1}{(\sqrt{2}+1)tan\frac{x}{2}}\right|{\int }_0^{\frac{\pi }{4}}$
$=\frac{1}{2}\sqrt{2}log\left(\right(\sqrt{2}+1{)}^2)=\frac{1}{\sqrt{2}}log(\sqrt{2}+1)$