Question
Evaluate: $\int\frac{\text{x sin}^{-1}\text{x}}{{\sqrt{\text{x-1}^{2}}}}\text{dx}.$
✓
Answer
$\text{I}=\int\frac{\text{x sin}^{-1}\text{x}}{{\sqrt{\text{1 - x}^{2}}}}\text{dx};$ Let x = sinθ $\Rightarrow$ dx = cosθ dθ
$=\int\frac{\sin\theta\cdot\theta}{\cos\theta}\cos\theta\text{ d}\theta=\int\theta\sin\theta\text{ d}\theta$
= –θ cosθ + $\int$cosθ dθ = –θ cosθ + sinθ + c
$=-\sin^{-1}\text{x}(\sqrt{\text{1-x}^{2}})+\text{x + c}$.
$=\int\frac{\sin\theta\cdot\theta}{\cos\theta}\cos\theta\text{ d}\theta=\int\theta\sin\theta\text{ d}\theta$
= –θ cosθ + $\int$cosθ dθ = –θ cosθ + sinθ + c
$=-\sin^{-1}\text{x}(\sqrt{\text{1-x}^{2}})+\text{x + c}$.
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