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Question

Question: \(\int _ { 0 } ^ { \pi / 2 } \frac { 1 } { 1 + \sqrt { \tan x } } d x =\)...

∫0π/211+tan⁡xdx=\int _ { 0 } ^ { \pi / 2 } \frac { 1 } { 1 + \sqrt { \tan x } } d x =

A

π2\frac { \pi } { 2 }

B

π4\frac { \pi } { 4 } =

C

π6\frac { \pi } { 6 }

D

1

Answer

π4\frac { \pi } { 4 } =

Explanation

Solution

∫0π/211+tan⁡xdx=∫0π/2cos⁡xcos⁡x+sin⁡xdx=π4\int _ { 0 } ^ { \pi / 2 } \frac { 1 } { 1 + \sqrt { \tan x } } d x = \int _ { 0 } ^ { \pi / 2 } \frac { \sqrt { \cos x } } { \sqrt { \cos x } + \sqrt { \sin x } } d x = \frac { \pi } { 4 } .