- xsin -1x dx- 1 - k - 1- x 2 x- px 2 -1 sin -1 x -Find k-p -

I=∫ x sin ^ 1 x dx #160;= 1 / 2 x ^ 2 sin ^ 1 x 1 / 2 ∫ x ^ 2 /√( 1 x ^ 2 ) #160; dx. Substitute #160;x=sin u ⇒ dx=cos u #160;du ...

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Byju's Answer

Standard XII

Mathematics

Theorems on Integration

∫ xsin -1x dx...

Question

$\int x {sin}^{- 1} x d x = \frac{1}{k} (\sqrt{1 - x^{2}} x + ({p x}^{2} - 1) {sin}^{- 1} x)$ .Find $k + p$ ?

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Solution

$I = \int x {sin}^{- 1} x d x$
$= \frac{1}{2} x^{2} {sin}^{- 1} x \frac{- 1}{2} \int \frac{x^{2}}{\sqrt{1 - x^{2}}} d x .$
Substitute $x = sin u \Rightarrow d x = cos u d u$
$I = \frac{1}{2} x^{2} {sin}^{- 1} x - \frac{1}{2} \int {sin}^{2} u d u$
$= \frac{1}{2} x^{2} {sin}^{- 1} x + \frac{1}{4} \int cos 2 u d u - \frac{1}{4} \int d u$
$= \frac{1}{2} x^{2} {sin}^{- 1} x + \frac{sin 2 u}{8} - \frac{u}{4}$
$= \frac{1}{4} (\sqrt{1 - x^{2}} x + ({2 x}^{2} - 1) {sin}^{- 1} x)$
$∴ k + p = 6$

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∫xsin−1xdx=1k(√1−x2x+(px2−1)sin−1x).Find k+p ?

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