# Evaluate $$\int{(sin^{-1}x)^2 dx}$$

Solution:

Let I = $$\int{(sin^{-1}x)^2 \enspace . \enspace 1 \enspace dx}$$

Taking (sin-1x)2 as the first function and 1 as second function and integrating by parts, we obtain

$$= (sin^{-1}x)^2 \int 1dx – \int\left \{ \frac{d}{dx} (sin^{-1}x)^2 . \int {1} \right \} dx$$

$$= (sin^{-1}x)^2 . x – \int \frac{2sin^{-1}x}{\sqrt{1 – x^2}} . xdx$$

$$= x(sin^{-1}x)^2 + \int sin^{-1}x . \left ( \frac{-2x}{\sqrt{1 – x^2}} \right )dx$$

$$= x(sin^{-1}x)^2 + \left [ sin^{-1}x \int\frac{-2x}{\sqrt{1-x^2}} dx – \int \left \{ \left ( \frac{d}{dx} sin^{-1}x \right ) \int \frac{-2x}{\sqrt{1-x^2}}dx \right \} dx \right ]$$

$$= x(sin^{-1}x)^2 + \left [ sin^{-1}.2\sqrt{1-x^2} -\int {\frac{1}{\sqrt{1-x^2}} . 2\sqrt{1-x^2}dx} \right ]$$

$$= x(sin^{-1}x)^2 + 2\sqrt{1-x^2} sin^{-1}x – \int 2dx$$

$$= x(sin^{-1}x)^2 + 2\sqrt{1-x^2} sin^{-1}x – 2x + C$$