Solveeit Logo
Login

Question

Question: The value of $\int \frac{dx}{\sqrt{(1-x^2)} \sin^{-1} x}$ is...

The value of ∫dx(1−x2)sin⁡−1x\int \frac{dx}{\sqrt{(1-x^2)} \sin^{-1} x} is

A

(sin⁡−1x)2+c(\sin^{-1}x)^2 + c

B

sin⁡−1x+c\sqrt{\sin^{-1}x} + c

C

2sin⁡−1x+c2\sqrt{\sin^{-1}x} + c

D

2(sin⁡−1x)2+c2(\sin^{-1}x)^2 + c

Answer

2\sqrt{\sin^{-1}x} + c

Explanation

Solution

The integral is ∫dx(1−x2)sin⁡−1x\int \frac{dx}{\sqrt{(1-x^2)} \sin^{-1} x}. Assuming the question intends the sin⁡−1x\sin^{-1} x term to be under the square root in the denominator, the integral is ∫dx1−x2sin⁡−1x\int \frac{dx}{\sqrt{1-x^2} \sqrt{\sin^{-1} x}}. Let u=sin⁡−1xu = \sin^{-1} x. Then du=11−x2dxdu = \frac{1}{\sqrt{1-x^2}} dx. The integral transforms to ∫1udu=∫u−1/2du\int \frac{1}{\sqrt{u}} du = \int u^{-1/2} du. Integrating u−1/2u^{-1/2} gives u1/21/2+C=2u+C\frac{u^{1/2}}{1/2} + C = 2\sqrt{u} + C. Substituting back u=sin⁡−1xu = \sin^{-1} x, we get 2sin⁡−1x+C2\sqrt{\sin^{-1} x} + C.