What is the integral of arcsin(x)?

integral of arcsin(x)

The integral of

\sin^{-1}(x)

x\sin^{-1}(x) + \sqrt{1 - x^2} + C

.

Solution. We want to find the integral of

\sin^{-1}(x)

, i.e.:

\begin{align*} \int \sin^{-1}(x) dx. \end{align*}

Firstly, we will apply the method integrating by parts, i.e.:

\begin{align*} \int UdV = UV - \int VdU, \end{align*}

where we get the following functions:

\begin{align*} U = \sin^{-1}(x), \quad &dV = dx\\ dU = \frac{dx}{\sqrt{1 - x^2}}, \quad &V = x. \end{align*}

The derivative

dU

can be easily verified here. Now we get the following integral:

\begin{align*} \int \sin^{-1}(x) dx = x\sin^{-1}(x) - \int \frac{x}{\sqrt{1 - x^2}} dx. \end{align*}

Secondly and lastly, we will use the substitution method. Let

u = 1 - x^2

, then

du = -2x dx

. We get the following:

\begin{align*} \int \sin^{-1}(x) dx &= x\sin^{-1}(x) - \int \frac{x}{\sqrt{1 - x^2}} dx \\ &= x\sin^{-1}(x) + \frac{1}{2} \int u^{-\frac{1}{2}} du \\ &= x\sin^{-1}(x) + u^{\frac{1}{2}} + C \\ &= x\sin^{-1}(x) + \sqrt{1 - x^2} + C. \end{align*}

Therefore, the integral of

\sin^{-1}(x)

x\sin^{-1}(x) + \sqrt{1 - x^2} + C