# Question 117a3

Jan 19, 2017

The integral is $\ln | \frac{1 + \sqrt{1 - {x}^{2}}}{x} | - {\cos}^{-} 1 \frac{x}{x} + C$

#### Explanation:

This can be rewritten as $\int {\cos}^{-} 1 x {x}^{-} 2 \mathrm{dx}$. Whenever you see a product within an integral, you should consider using integration by parts.

The integration by parts formula states that an integral $\int u \mathrm{dv} = u v - \int \left(v \mathrm{du}\right)$. ${\cos}^{-} 1 x$ is not that simple to integrate, so let's say $u = {\cos}^{-} 1 x$ and $\mathrm{dv} = {x}^{-} 2$. Then $\mathrm{du} = - \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$ and $v = - \frac{1}{x}$.

Use the formula now:

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

$\int {\cos}^{-} 1 x {x}^{-} 2 \mathrm{dx} = - {\cos}^{-} 1 \frac{x}{x} - \int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx}$

The remaining integral will have to be integrated by trigonometric substitution. This is a technique of integration where you make use of a pythagorean identity to get rid of the √#. When the square root within the integral is of the form $\sqrt{{a}^{2} - {x}^{2}}$ like the one above, use the substitution $\textcolor{red}{x = \sin \theta}$. Then $\mathrm{dx} = \cos \theta d \theta$.

$\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = \int \frac{1}{\sin \theta \sqrt{1 - {\sin}^{2} \theta}} \cdot \cos \theta d \theta$

$\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = \int \frac{1}{\sin \theta \sqrt{{\cos}^{2} \theta}} \cdot \cos \theta d \theta$

$\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = \int \frac{1}{\sin} \theta d \theta$

$\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = \int \csc \theta d \theta$

This is a known integral that is derived here.

$\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = - \ln | \csc \theta + \cot \theta | + C$

We now reverse the substitutions by drawing a triangle and finding the correct ratios. We know from our original substitution that $\frac{x}{1} = \sin \theta$. $\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = - \ln | \frac{1}{x} + \frac{\sqrt{1 - {x}^{2}}}{x} | + C$

$\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = - \ln | \frac{1 + \sqrt{1 - {x}^{2}}}{x} | + C$

Let's return our attention to the whole integral.

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

$\int {\cos}^{-} 1 x {x}^{-} 2 \mathrm{dx} = \ln | \frac{1 + \sqrt{1 - {x}^{2}}}{x} | - {\cos}^{-} 1 \frac{x}{x} + C$

Hopefully this helps!