Question

$\int _ { \pi / 6 } ^ { \pi / 3 } \frac { d x } { 1 + \sqrt { \cot x } }$ is

• A. $\pi / 3$
• B. $\pi / 6$
• C. $\pi / 12$
• D. $\pi / 2$

Answer 1

Answer: (C)

$\pi / 12$

Answer 2

$I = \int _ { \pi / 6 } ^ { \pi / 3 } \frac { d x } { 1 + \sqrt { \cot x } } = \int _ { \pi / 6 } ^ { \pi / 3 } \frac { \sqrt { \sin x } } { \sqrt { \sin x } + \sqrt { \cos x } } d x$ …..(i)

$I = \int _ { \pi / 6 } ^ { \pi / 3 } \frac { \sqrt { \cos x } } { \sqrt { \cos x } + \sqrt { \sin x } } d x$ .....(ii)

Adding (i) and (ii),

$2 I = \int _ { \pi / 6 } ^ { \pi / 3 } d x$ ; $I = \frac { 1 } { 2 } \left( \frac { \pi } { 3 } - \frac { \pi } { 6 } \right) = \frac { \pi } { 12 }$.