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Question: \(\int _ { 0 } ^ { 2 } \sqrt { \frac { 2 + x } { 2 - x } } d x =\)...

∫022+x2−xdx=\int _ { 0 } ^ { 2 } \sqrt { \frac { 2 + x } { 2 - x } } d x =

A

π+2\pi + 2

B

π+32\pi + \frac { 3 } { 2 }

C

π+1\pi + 1

D

None of these

Answer

π+2\pi + 2

Explanation

Solution

Put x=2cos⁡θ⇒dx=−2sin⁡θdθx = 2 \cos \theta \Rightarrow d x = - 2 \sin \theta d \thetathen

∫022+x2−xdx=−2∫π/201+cos⁡θ1−cos⁡θsin⁡θdθ\int _ { 0 } ^ { 2 } \sqrt { \frac { 2 + x } { 2 - x } } d x = - 2 \int _ { \pi / 2 } ^ { 0 } \sqrt { \frac { 1 + \cos \theta } { 1 - \cos \theta } } \sin \theta d \theta

=4∫0π/2cos⁡(θ/2)sin⁡(θ/2)sin⁡θ2cos⁡θ2dθ= 4 \int _ { 0 } ^ { \pi / 2 } \frac { \cos ( \theta / 2 ) } { \sin ( \theta / 2 ) } \sin \frac { \theta } { 2 } \cos \frac { \theta } { 2 } d \theta

=2∫0π/2(1+cos⁡θ)dθ= 2 \int _ { 0 } ^ { \pi / 2 } ( 1 + \cos \theta ) d \theta

=2[θ+sin⁡θ]0π/2=2[π2+1]=π+2= 2 [ \theta + \sin \theta ] _ { 0 } ^ { \pi / 2 } = 2 \left[ \frac { \pi } { 2 } + 1 \right] = \pi + 2 .