What is the integral of arccos(x)?

integral of arccos(x)

The integral of

\cos^{-1}(x)

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

.

Solution. We want to determine the integral of

\cos^{-1}(x)

, i.e.:

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

The firs step we will take is integrating by parts, i.e.:

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

where we get the following functions:

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

We have already proved earlier the derivative

dU

here. We get the next integral:

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

Lastly, we will use the substitution method. Let

u = 1 - x^2

, then

du = -2x dx

. We get the following:

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

Therefore, the integral of

\cos^{-1}(x)

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