How do you find the integral of (sin^(-1)x)^2?

Jan 23, 2017

$\int {\left({\sin}^{-} 1 x\right)}^{2} \mathrm{dx} = x {\left({\sin}^{-} 1 x\right)}^{2} + 2 \sqrt{1 - {x}^{2}} {\sin}^{-} 1 x - 2 x + C$

Explanation:

$I = \int {\left({\sin}^{-} 1 x\right)}^{2} \mathrm{dx}$

We will use integration by parts. This takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. For the given integral $I$, let:

$\left\{\begin{matrix}u = {\left({\sin}^{-} 1 x\right)}^{2} \\ \mathrm{dv} = \mathrm{dx}\end{matrix}\right.$

Differentiating $u$ and integrating $\mathrm{dv}$ show:

$\left\{\begin{matrix}\mathrm{du} = \frac{2 {\sin}^{-} 1 x}{\sqrt{1 - {x}^{2}}} \mathrm{dx} \\ v = x\end{matrix}\right.$

Note that differentiating $u$ requires the chain rule.

Then:

$I = u v - \int v \mathrm{du} = x {\left({\sin}^{-} 1 x\right)}^{2} - \int \frac{2 x \left({\sin}^{-} 1 x\right)}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

Perform integration by parts again. Note that the portion $\int \frac{2 x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$ is easily integrable with the substitution $t = 1 - {x}^{2} \implies \mathrm{dt} = - 2 x \mathrm{dx}$, so:

$\int \frac{2 x}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = - \int {t}^{- \frac{1}{2}} \mathrm{dt} = - 2 \sqrt{t} = - 2 \sqrt{1 - {x}^{2}}$

Then for integration by parts let:

$\left\{\begin{matrix}u = {\sin}^{-} 1 x \text{ "=>" "du=1/sqrt(1-x^2)dx \\ dv=(2x)/sqrt(1-x^2)dx" "=>" } v = - 2 \sqrt{1 - {x}^{2}}\end{matrix}\right.$

So:

$I = x {\left({\sin}^{-} 1 x\right)}^{2} - \left(u v - \int v \mathrm{du}\right)$

$I = x {\left({\sin}^{-} 1 x\right)}^{2} - u v + \int v \mathrm{du}$

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

$I = x {\left({\sin}^{-} 1 x\right)}^{2} + 2 \sqrt{1 - {x}^{2}} {\sin}^{-} 1 x - 2 \int \mathrm{dx}$

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