Solveeit Logo
Login

Question

Question: The value of \(\int _ { 0 } ^ { 1 } \frac { d x } { x + \sqrt { 1 - x ^ { 2 } } }\) is...

The value of ∫01dxx+1−x2\int _ { 0 } ^ { 1 } \frac { d x } { x + \sqrt { 1 - x ^ { 2 } } } is

A

π3\frac { \pi } { 3 }

B

π2\frac { \pi } { 2 }

C

12\frac { 1 } { 2 }

D

π4\frac { \pi } { 4 }

Answer

π4\frac { \pi } { 4 }

Explanation

Solution

∫01dxx+1−x2=∫0π/2cos⁡θdθsin⁡θ+cos⁡θ\int _ { 0 } ^ { 1 } \frac { d x } { x + \sqrt { 1 - x ^ { 2 } } } = \int _ { 0 } ^ { \pi / 2 } \frac { \cos \theta d \theta } { \sin \theta + \cos \theta } =π4= \frac { \pi } { 4 },

(Put x=sin⁡θ,dx=cos⁡θdθx = \sin \theta , d x = \cos \theta d \theta).