Evaluate 1 . sin - 1 x ? d x .

Question

Evaluate $\int 1 . \sin^{- 1} x d x$ .

Smart Achievers · Accepted Answer

(4)

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

Explanation :

Let, the given integral be

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

Integrating by parts,

$I = \sin^{- 1} x \int 1 . d x - \int {(\frac{d}{d x} \sin^{- 1} x) \int 1 . d x} d x$

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

Substitute $1 - x^{2} = t$ in the given integral and differentiate it.

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

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

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

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

$= x \sin^{- 1} x + \frac{1}{2} t^{\frac{1}{2}} \times (2) + C$

$= x \sin^{- 1} x + {(1 - x^{2})}^{\frac{1}{2}} + C$ $(∵ t = 1 - x^{2})$

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