∫ x sin^ 1x dx Let x=sin θ dx=cosθ dθ Substituting values, we get ∫ x sin^ 1x dx=∫sinθ sin^ 1(sinθ)cosθ dθ #160; #160; #160; #160 ...

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Integration by Parts

∫xsin - 1xdx

Question

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

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Solution

$\int x {sin}^{- 1} x d x$
Let $x = sin θ$
$d x = cos θ d θ$
Substituting values, we get
$\int x {sin}^{- 1} x d x = \int sin θ {sin}^{- 1} (sin θ) cos θ d θ$

$= \int sin θ θ cos θ d θ$
[ Since ${sin}^{- 1} (sin θ) = θ$ ]

$= \frac{1}{2} \int θ (2 sin θ cos θ) d θ$

$= \frac{1}{2} \int θ sin 2 θ d θ$ [ Since, $2 sin x cos x = sin 2 x$ ]

$= \frac{1}{2} [θ \int sin 2 θ d θ - \int (\frac{d θ}{d θ} \int sin 2 θ) d θ]$

$= \frac{1}{2} [θ \frac{- (cos 2 θ)}{2} - \int 1. \frac{- (cos 2 θ)}{2} d θ]$

$= \frac{1}{2} [\frac{- θ}{2} cos 2 θ + \frac{1}{2} \int cos 2 θ d θ]$

$= \frac{1}{2} [\frac{- θ}{2} cos 2 θ + \frac{1}{2} \frac{sin 2 θ}{2}] + c$

$= \frac{- θ}{4} cos 2 θ + \frac{1}{8} sin 2 θ + c$

$= \frac{- θ}{4} (1 - 2 {sin}^{2} θ) + \frac{1}{8} \times 2 sin θ cos θ + c$

$= \frac{- θ}{4} (1 - 2 {sin}^{2} θ) + \frac{1}{4} sin θ cos θ + c$

$= \frac{- θ}{4} (1 - 2 {sin}^{2} θ) + \frac{1}{4} sin θ \sqrt{1 - {sin}^{2} θ} + c$

$= \frac{- {sin}^{- 1} x}{4} (1 - 2 x^{2}) + \frac{x}{4} \sqrt{1 - x^{2}} + c$ [ Since, $θ = {sin}^{- 1} x$ ]

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

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