# Question #77e15

Feb 21, 2016

$\setminus \frac{1}{3} {x}^{3} \setminus {\sin}^{-} 1 \left(x\right) + \setminus \frac{1}{3} \left(\setminus \sqrt{1 - {x}^{2}} - \setminus \frac{1}{3} {\left(1 - {x}^{2}\right)}^{\setminus \frac{3}{2}}\right) + C$

#### Explanation:

$\setminus \int {x}^{2} {\sin}^{- 1} x \mathrm{dx}$

Applying integration by parts:
=$\int \setminus u v ' = u v - \setminus \int u ' v$ ...(i)

=$u = \setminus {\sin}^{-} 1 \left(x\right) , u ' = \setminus \frac{1}{\setminus \sqrt{1 - {x}^{2}}} , v ' = {x}^{2} , v = \setminus \frac{{x}^{3}}{3}$

=$u = \setminus {\sin}^{-} 1 \left(x\right)$ $v ' = {x}^{2}$

=$u ' = \setminus \frac{d}{\mathrm{dx}} \left(\setminus {\sin}^{-} 1 \left(x\right)\right)$ = $\setminus \frac{1}{\setminus \sqrt{1 - {x}^{2}}}$

(Applying common derivative $\setminus \frac{d}{\mathrm{dx}} \left(\setminus {\sin}^{-} 1 \left(x\right)\right) = \setminus \frac{1}{\setminus \sqrt{1 - {x}^{2}}}$

=$v = \setminus \int {x}^{2} \mathrm{dx}$ = ${x}^{3} / 3$

=$u = \setminus {\sin}^{-} 1 \left(x\right) , u ' = \setminus \frac{1}{\setminus \sqrt{1 - {x}^{2}}} , v ' = {x}^{2} , v = \setminus \frac{{x}^{3}}{3}$

=${\sin}^{-} 1 \left(x\right) \setminus \frac{{x}^{3}}{3} - \setminus \int \setminus \frac{1}{\setminus \sqrt{1 - {x}^{2}}} \setminus \frac{{x}^{3}}{3} \mathrm{dx}$ (from i)

=$\setminus \frac{1}{3} {x}^{3} \setminus {\sin}^{-} 1 \left(x\right) - \setminus \int \setminus \frac{{x}^{3}}{3 \setminus \sqrt{1 - {x}^{2}}} \mathrm{dx}$

We have $\setminus \int \setminus \frac{{x}^{3}}{3 \setminus \sqrt{1 - {x}^{2}}} \mathrm{dx}$ = $\setminus \frac{1}{3} \left(\setminus \frac{1}{3} {\left(1 - {x}^{2}\right)}^{\setminus \frac{3}{2}} - \setminus \sqrt{1 - {x}^{2}}\right)$

(Taking the constant out $\setminus \int a \setminus \cdot f \left(x\right) \mathrm{dx} = a \setminus \cdot \setminus \int f \left(x\right) \mathrm{dx}$ and applying sum rule $\setminus \int f \left(x\right) \setminus \pm g \left(x \setminus\right) \mathrm{dx} = \setminus \int f \left(x\right) \mathrm{dx} \setminus \pm \setminus \int g \left(x\right) \mathrm{dx}$)

=$\setminus \frac{1}{3} {x}^{3} \setminus {\sin}^{-} 1 \left(x\right)$ - $\setminus \frac{1}{3} \left(\setminus \frac{1}{3} {\left(1 - {x}^{2}\right)}^{\setminus \frac{3}{2}} - \setminus \sqrt{1 - {x}^{2}}\right)$

Simplifying it
=$\setminus \frac{1}{3} {x}^{3} {\sin}^{-} 1 \left(x\right) + \setminus \frac{1}{3} \left(\setminus \sqrt{1 - {x}^{2}} - \setminus \frac{1}{3} {\left(1 - {x}^{2}\right)}^{\setminus \frac{3}{2}}\right)$

=$\setminus \frac{1}{3} {x}^{3} \setminus {\sin}^{-} 1 \left(x\right) + \setminus \frac{1}{3} \left(\setminus \sqrt{1 - {x}^{2}} - \setminus \frac{1}{3} {\left(1 - {x}^{2}\right)}^{\setminus \frac{3}{2}}\right) + C$