arccos(x) Archives - Epsilonify

What is the integral of arccos(x)?

The Mathematician — Wed, 12 Apr 2023 13:00:33 +0000

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

The Mathematician — Sat, 03 Dec 2022 13:00:05 +0000

The derivative of

\arccos(x)

-\frac{1}{\sqrt{1 - x^2}}

.

Solution. Let

y = \cos^{-1}(x)

. Then

x = \cos(y)

and

0 \leq y \leq \pi

. We want to differentiate with respect to

x

\begin{align*} \frac{d}{dx} x = \frac{d}{dx} \cos(y) &\iff 1 = \frac{d(\cos(y))}{dy}\frac{dy}{dx} \\ &\iff 1 = -\sin(y) \frac{dy}{dx} \\ &\iff \frac{dy}{dx} = -\frac{1}{\sin(y)}, \end{align*}

where we have seen here that

\frac{d(\cos(y))}{dy} = -\sin(y)

. We have that

0 \leq y \leq \pi

and therefore

\sin(y) \geq 1

. So we can apply the next identity:

\begin{align*} \sin^2(y) + \cos^2(y) = 1 &\iff \sin^2(y) = 1 - \cos^2(y) \\ &\iff \sin(y) = \sqrt{1 - \cos^2(y)} \\ &\iff \sin(y) = \sqrt{1 - x^2} \end{align*}

where

\cos^2(y) = x^2

since

\cos(y) = x

. Combining everything, we get that the derivative of

\arccos(x)

is:

\begin{align*} \frac{d}{dx} \arccos(x) = \frac{d}{dx} \cos^{-1}(x) = \frac{dy}{dx} = -\frac{1}{\sin(y)} = -\frac{1}{\sqrt{1-x^2}}, \quad x \in (-1,1). \end{align*}

So, the derivative of

\arccos(x)

-\frac{1}{\sqrt{1-x^2}}