First slide

Introduction to integration

Question

$\int {xsin}^{- 1} xdx$ is equal to

Moderate

A

$\frac{\sin^{- 1} x}{2} (2 x^{2} - 1) + \frac{1}{4} x \sqrt{1 - x^{2}} + C$
B

$\frac{\sin^{- 1} x}{4} (2 x^{2} - 1) + \frac{1}{4} x \sqrt{1 - x^{2}} + C$
C

$\frac{\sin^{- 1} x}{4} (2 x^{2} - 1) + \frac{1}{2} x \sqrt{1 - x^{2}} + C$
D

None of the above

Solution

Let $I = \int x \sin^{- 1} xdx$
On taking sin^-1 x as first function and x as second function and integrating by parts, we get
[ $∵$ inverse functions comes before,algebraic functions in ILATE]
$\begin{array}{l} I = \sin^{- 1} x \int xdx - \int [\frac{d}{dx} (\sin^{- 1} x) \int xdx] dx \\ = \sin^{- 1} x \cdot \frac{x^{2}}{2} - \int [\frac{1}{\sqrt{1 - x^{2}}} \cdot \frac{x^{2}}{2}] dx \end{array}$
$= \frac{x^{2}}{2} \cdot \sin^{- 1} x - \int [\frac{1 - (1 - x^{2})}{\sqrt{1 - x^{2}}} \cdot \frac{1}{2}] dx$
[add and subtract 1 in numerator of second term]
$\begin{array}{l} = \frac{x^{2}}{2} \cdot \sin^{- 1} x - \frac{1}{2} \int \frac{1}{\sqrt{1 - x^{2}}} dx + \frac{1}{2} \int \frac{1 - x^{2}}{\sqrt{1 - x^{2}}} dx \\ = \frac{x^{2}}{2} \cdot \sin^{- 1} x - \frac{1}{2} \sin^{- 1} x + \frac{1}{2} \int \sqrt{1 - x^{2}} dx \end{array}$
$\begin{array}{l} \Rightarrow I = \frac{x^{2}}{2} \cdot \sin^{- 1} x - \frac{1}{2} \sin^{- 1} x + \frac{1}{2} [\frac{1}{2} x \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{- 1} x] + C \\ \Rightarrow I = \sin^{- 1} x [\frac{x^{2}}{2} - \frac{1}{4}] + \frac{1}{4} x \sqrt{1 - x^{2}} + C \\ \Rightarrow I = \frac{\sin^{- 1} x}{4} (2 x^{2} - 1) + \frac{1}{4} x \sqrt{1 - x^{2}} + C \end{array}$

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