Question

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

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

Answer 1

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

$I={\int }_0^{{}^{\pi }{/}_2}\frac{sinx}{(sinx+cosx{)}^2}dx$ ---- (1)

$I={\int }_0^{{}^{\pi }{/}_2}\frac{cosx}{(sinx+cosx{)}^2}dx$ ---- (2)

$2I={\int }_0^{{}^{\pi }{/}_2}\frac{1}{sinx+cosx}dx$

$2I={\int }_0^{{}^{\pi }{/}_2}\frac{2dta{n}^x{/}_2}{2-(ta{n}^x{/}_2-1{)}^2}$

$I{=}^1{/}_2\sqrt{2}log\left|\frac{\sqrt{2}+(ta{n}^x{/}_2-1)}{\sqrt{2}-(ta{n}^x{/}_2-1)}\right|{|}_0^1$

${=}^1{/}_{\sqrt{2}}log(\sqrt{2}+1)$