Question

The value of the definite integral $\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{\sqrt{1+\sin x}}{\cos x}dx$ is equal to

• A. $\sqrt{2}\ln (\sqrt{8}-\sqrt{6}+\sqrt{4}-\sqrt{3})$
• B. $\sqrt{2}\ln (\sqrt{8}+\sqrt{6}-\sqrt{4}-\sqrt{3})$
• C. $\frac{1}{\sqrt{2}}\ln 2$
• D. $\sqrt{2}\ln 2$

Answer 1

Answer: (B)

$I=\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{\sqrt{1+\sin x}}{\cos x}dx=\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}dx=\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{dx}{\cos \frac{x}{2}-\sin \frac{x}{2}}=\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{dx}{\sqrt{2}\cos \left(\frac{x}{2}+\frac{\pi }{4}\right)}$
$=\frac{1}{\sqrt{2}}\underset{0}{\overset{\frac{\pi }{6}}{\int }}\sec \left(\frac{x}{2}+\frac{\pi }{4}\right)dx=\sqrt{2}{\left[\ln \left(\sec \left(\frac{\pi }{4}+\frac{x}{2}\right)+\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)\right)\right]}_0^{\frac{\pi }{6}}$
$=\sqrt{2}\ln \left(\frac{2+\sqrt{3}}{\sqrt{2}+1}\right)=\sqrt{2}\ln (\sqrt{8}+\sqrt{6}-\sqrt{4}-\sqrt{3})$