QuestionFind the integral of the inverse cosine function: $\int \cos^{-1} x \, dx$ .

Studdy Solution

STEP 1

Assumptions1. We know the integral of $\cos^{-1} x$ . . We know the integral and derivative properties.

STEP 2

We know that the integral of a function $f(x)$ is given by $(x) + C$ , where $'(x) = f(x)$ and $C$ is the constant of integration.

STEP 3

The integral of $\cos^{-1} x$ is not a standard integral that we can directly compute. However, we can use the method of integration by parts to solve it. The formula for integration by parts is $\int u dv = uv - \int v du$

STEP 4

We will choose $u = \cos^{-1} x$ and $dv = dx$ . Then we need to find $du$ and $v$ .

STEP 5

To find $du$ , we differentiate $u = \cos^{-1} x$ with respect to $x$ .
$du = -\frac{1}{\sqrt{1 - x^2}} dx$

STEP 6

To find $v$ , we integrate $dv = dx$ with respect to $x$ .
$v = \int dx = x$

STEP 7

Now we substitute $u$ , $v$ , $du$ , and $dv$ into the integration by parts formula.
$\int \cos^{-1} x dx = uv - \int v du$

STEP 8

Substitute the values of $u$ , $v$ , $du$ , and $dv$ .
$\int \cos^{-1} x dx = x \cos^{-1} x - \int x \left(-\frac{1}{\sqrt{1 - x^2}}\right) dx$

STEP 9

implify the integral.
$\int \cos^{-} x dx = x \cos^{-} x + \int \frac{x}{\sqrt{ - x^2}} dx$

STEP 10

The integral $\int \frac{x}{\sqrt{ - x^2}} dx$ is a standard integral, which can be solved by substituting $t = - x^2$ .

STEP 11

Differentiate t = - x^ with respect to $x$ to find $dt$ .
$dt = -x dx$

STEP 12

olve the equation $dt = -2x dx$ for $dx$ .
$dx = -\frac{dt}{2x}$

STEP 13

Substitute $t = - x^2$ and $dx = -\frac{dt}{2x}$ into the integral $\int \frac{x}{\sqrt{ - x^2}} dx$ .
$\int \frac{x}{\sqrt{ - x^2}} dx = \int \frac{x}{\sqrt{t}} \left(-\frac{dt}{2x}\right)$

STEP 14

implify the integral.
$\int \frac{x}{\sqrt{t}} \left(-\frac{dt}{2x}\right) = -\frac{}{2} \int \frac{dt}{\sqrt{t}}$

STEP 15

The integral $\int \frac{dt}{\sqrt{t}}$ is a standard integral, which is $2\sqrt{t}$ .
$-\frac{}{2} \int \frac{dt}{\sqrt{t}} = -\sqrt{t}$

STEP 16

Substitute $t = - x^2$ back into the integral.
$-\sqrt{t} = -\sqrt{ - x^2}$

STEP 17

Substitute $-\sqrt{ - x^2}$ back into the original integral.
$\int \cos^{-} x dx = x \cos^{-} x + \int \frac{x}{\sqrt{ - x^2}} dx = x \cos^{-} x - \sqrt{ - x^2}$

STEP 18

Finally, we add the constant of integration $C$ to the result.
$\int \cos^{-} x dx = x \cos^{-} x - \sqrt{ - x^2} + C$ So, the integral of $\cos^{-} x$ with respect to $x$ is $x \cos^{-} x - \sqrt{ - x^2} + C$ .

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