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Question

Question: \(\int _ { 0 } ^ { \pi } \frac { x \tan x } { \sec x + \tan x } d x =\)...

∫0πxtan⁡xsec⁡x+tan⁡xdx=\int _ { 0 } ^ { \pi } \frac { x \tan x } { \sec x + \tan x } d x =

A

π2−1\frac { \pi } { 2 } - 1

B

π(π2+1)\pi \left( \frac { \pi } { 2 } + 1 \right)

C

π2+1\frac { \pi } { 2 } + 1

D

π(π2−1)\pi \left( \frac { \pi } { 2 } - 1 \right)

Answer

π(π2−1)\pi \left( \frac { \pi } { 2 } - 1 \right)

Explanation

Solution

I=∫0πxtan⁡xsec⁡x+tan⁡xdx=∫0π(π−x)tan⁡(π−x)sec⁡(π−x)+tan⁡(π−x)dxI = \int _ { 0 } ^ { \pi } \frac { x \tan x } { \sec x + \tan x } d x = \int _ { 0 } ^ { \pi } \frac { ( \pi - x ) \tan ( \pi - x ) } { \sec ( \pi - x ) + \tan ( \pi - x ) } d x

⇒ 2I=π2∫0πtan⁡xsec⁡x+tan⁡xdx=π2∫0πsin⁡x1+sin⁡xdx2 I = \frac { \pi } { 2 } \int _ { 0 } ^ { \pi } \frac { \tan x } { \sec x + \tan x } d x = \frac { \pi } { 2 } \int _ { 0 } ^ { \pi } \frac { \sin x } { 1 + \sin x } d x

=π2[∫0π1dx−∫0πdx1+sin⁡x]\frac { \pi } { 2 } \left[ \int _ { 0 } ^ { \pi } 1 d x - \int _ { 0 } ^ { \pi } \frac { d x } { 1 + \sin x } \right]

On solving, we get I=π22−π=π(π2−1)I = \frac { \pi ^ { 2 } } { 2 } - \pi = \pi \left( \frac { \pi } { 2 } - 1 \right).