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Question: The value of \(\int _ { \pi / 4 } ^ { 3 \pi / 4 } \frac { \phi } { 1 + \sin \phi } d \phi\) is...

The value of ∫π/43π/4ϕ1+sin⁡ϕdϕ\int _ { \pi / 4 } ^ { 3 \pi / 4 } \frac { \phi } { 1 + \sin \phi } d \phi is

A

πtan⁡π8\pi \tan \frac { \pi } { 8 }

B

C

tan⁡π8\tan \frac { \pi } { 8 }

D

None of these

Answer

πtan⁡π8\pi \tan \frac { \pi } { 8 }

Explanation

Solution

I=∫π/43π/4ϕ1+sin⁡ϕdϕ=∫π/43π/4π−ϕ1+sin⁡(π−ϕ)dϕI = \int _ { \pi / 4 } ^ { 3 \pi / 4 } \frac { \phi } { 1 + \sin \phi } d \phi = \int _ { \pi / 4 } ^ { 3 \pi / 4 } \frac { \pi - \phi } { 1 + \sin ( \pi - \phi ) } d \phi

{∵π4+3π4=π}\left\{ \because \frac { \pi } { 4 } + \frac { 3 \pi } { 4 } = \pi \right\}

⇒ 2I=∫π/43π/4π1+sin⁡ϕdϕ2 I = \int _ { \pi / 4 } ^ { 3 \pi / 4 } \frac { \pi } { 1 + \sin \phi } d \phi

On simplification, we get

I=π(2−1)=πtan⁡π8I = \pi ( \sqrt { 2 } - 1 ) = \pi \tan \frac { \pi } { 8 }