Question

${\int }_0^{\pi /2}\frac{1}{\sin x+\cos x}dx$

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

Answer 1

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

${\int }_0^{\frac{\pi }{2}}\frac{1}{sinx+cosx}dx$

${\int }_0^{\frac{\pi }{2}}\frac{1+ta{n}^2\frac{x}{2}}{1-ta{n}^2\frac{x}{2}+2tan\frac{x}{2}}dx$

${\int }_0^{\frac{\pi }{2}}\frac{2d\left(tan\frac{x}{2}\right)}{1-(tan\frac{x}{2}+1{)}^2}=2{\int }_0^{\frac{\pi }{2}}\frac{d\left(tan\frac{x}{2}\right)}{(\sqrt{2}{)}^2-(tan\frac{x}{2}-1{)}^2}$

$=2\left[\frac{1}{2\sqrt{2}}\right]log{\left|\frac{\sqrt{2}+tan\frac{x}{2}-1}{\sqrt{2}-(tan\frac{x}{2}-1)}\right|}_0^{\frac{\pi }{2}}$

$=\frac{1}{\sqrt{2}}log\left(\frac{\sqrt{2}+2}{\sqrt{2}-2}\right)$

$=log\left(\frac{1+\sqrt{2}}{1-\sqrt{2}}\right)$

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