Find the integral of x 2 . sin - 1 x ? d x

Question

Find the integral of $x^{2} . \sin^{- 1} x d x$

Smart Achievers · Accepted Answer

(2)

Ans. $\frac{1}{3} x^{3} \sin^{- 1} x + \frac{1}{3} \sqrt{1 - x^{2}} - \frac{1}{9} {(1 - x^{2})}^{3 / 2} + C$

Explanation:

The given integral is

$I = \int x^{2} . \sin^{- 1} x d x$

Integrating by parts, taking $\sin^{- 1} x$ as the first function, we get

$I = \int x^{2} . \sin^{- 1} x d x$

$= \sin^{- 1} x \cdot \frac{x^{3}}{3} - \int \frac{1}{\sqrt{1 - x^{2}}} \cdot \frac{x^{3}}{3} d x$

$= \sin^{- 1} x \cdot \frac{x^{3}}{3} - \frac{1}{3} \int \frac{x^{3}}{\sqrt{1 - x^{2}}} d x$

$= \frac{x^{3}}{3} . \sin^{- 1} x - \frac{1}{3} \int \frac{x . x^{2}}{\sqrt{1 - x^{2}}} d x$

Substituting $t^{2}$ for $1 - x^{2}$ and differentiate w.r.t. $x$ .

$\frac{d}{d x} (1 - x^{2}) = \frac{d}{d x} (t^{2})$

$\Rightarrow$ $- 2 x = 2 t \frac{d t}{d x}$

$\Rightarrow$ $x d x = - t d t$

Substitute the values, $1 - x^{2} = t^{2}$ and $x d x = - t d t$ in the integral.

$∴$ $I = \frac{x^{3}}{3} \sin^{- 1} x - \frac{1}{3} \int \frac{(- t) (1 - t^{2})}{t} d t$

$= \frac{x^{3}}{3} \sin^{- 1} x + \frac{1}{3} \int 1 d t - \frac{1}{3} \int t^{2} d t$

$= \frac{x^{3}}{3} \sin^{- 1} x + \frac{t}{3} - \frac{t^{3}}{9} + C$

$= \frac{1}{3} x^{3} \sin^{- 1} x + \frac{1}{3} \sqrt{1 - x^{2}} - \frac{1}{9} {(1 - x^{2})}^{3 / 2} + C$

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