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

∫0π/2dθ1+tan⁡θ=\int _ { 0 } ^ { \pi / 2 } \frac { d \theta } { 1 + \tan \theta } =

A

π\pi

B

π2\frac { \pi } { 2 }

C

π3\frac { \pi } { 3 }

D

π4\frac { \pi } { 4 }

Answer

π4\frac { \pi } { 4 }

Explanation

Solution

I=∫0π/2dθ1+tan⁡θ=∫0π/2dθ1+tan⁡(π2−θ)I = \int _ { 0 } ^ { \pi / 2 } \frac { d \theta } { 1 + \tan \theta } = \int _ { 0 } ^ { \pi / 2 } \frac { d \theta } { 1 + \tan \left( \frac { \pi } { 2 } - \theta \right) } =∫0π/2dθ1+cot⁡θ= \int _ { 0 } ^ { \pi / 2 } \frac { d \theta } { 1 + \cot \theta }

On adding, 2I=∫0π/2(11+tan⁡θ+11+cot⁡θ)dθ2 I = \int _ { 0 } ^ { \pi / 2 } \left( \frac { 1 } { 1 + \tan \theta } + \frac { 1 } { 1 + \cot \theta } \right) d \theta

=∫0π/2dθ=[θ]0π/2=π2⇒I=π4\int _ { 0 } ^ { \pi / 2 } d \theta = [ \theta ] _ { 0 } ^ { \pi / 2 } = \frac { \pi } { 2 } \Rightarrow I = \frac { \pi } { 4 }.