`int_(0)^(pi//2)(1)/((1+tanx))dx=?`

Options

A. `pi`
B. `pi//2`
C. `pi//3`
D. `pi//4`

Answer

D

Explanation

Given, `I=int _(0)^(pi//2)(dx)/(1+tan x)`
`I=int _(0)^(pi//2)(cos x)/(sin x+cos x)dx" "(i)`
`I=int _(0)^(pi//2)(cos (pi//2-x))/(sin (pi//2-x)+cos (pi//2-x))dx`
`I=int _(0)^(pi//2)(sin x)/(cos x+sin x)dx" "...(ii)`
On adding Eqs. (i) and (ii), we get
`2I=int _(0)^(pi//2)((sin x+cos x)/(sin x+cos x))dx`
`=int _(0)^(pi//2)dx=pi//2`
`impliesI=pi//4`