# How do you integrate int (sin^-1x)^2 using integration by parts?

Dec 17, 2016

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

#### Explanation:

We have:

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

Integration by parts takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. For $\int {\left({\sin}^{-} 1 x\right)}^{2} \mathrm{dx}$, we have to choose values of $u$ and $\mathrm{dv}$.

Whatever value of $\mathrm{dv}$ has to be integrated, so it would be foolish to choose ${\sin}^{-} 1 x$ or ${\left({\sin}^{-} 1 x\right)}^{2}$ as $\mathrm{dv}$ because their integrals are not clear. So, let $u = {\left({\sin}^{-} 1 x\right)}^{2}$ and $\mathrm{dv} = \mathrm{dx}$, all that remains.

So, we have:

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

Take the derivative of $u$ and integrate $\mathrm{dv}$:

$\left\{\begin{matrix}u = {\left({\sin}^{-} 1 x\right)}^{2} & \implies & \mathrm{du} = \frac{2 {\sin}^{-} 1 x}{\sqrt{1 - {x}^{2}}} \mathrm{dx} \\ \mathrm{dv} = \mathrm{dx} & \implies & v = x\end{matrix}\right.$

So we see that:

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

Or:

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

We now have another integral to use integration by parts on. Again, don't choose ${\sin}^{-} 1 x$ as $\mathrm{dv}$, so let $\mathrm{dv}$ be everything else.

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

Now differentiate and integrate, respectively. Note that $\int \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$ can be performed with the substitution $t = 1 - {x}^{2}$.

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

Then:

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

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

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

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

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