How do you evaluate the integral int x^2arcsecx?

Jun 22, 2017

$\frac{1}{3} {x}^{3} {\sec}^{-} 1 \left(x\right) - \frac{1}{6} x \sqrt{{x}^{2} - 1} - \frac{1}{6} \ln \left\mid x + \sqrt{{x}^{2} - 1} \right\mid + C$

Explanation:

$I = \int {x}^{2} {\sec}^{-} 1 \left(x\right) \mathrm{dx}$

Let $x = \sec \theta$. Then, $\theta = {\sec}^{-} 1 \left(x\right)$ and $\mathrm{dx} = \sec \theta \tan \theta d \theta$.

$I = \int {\sec}^{2} \theta \left(\theta\right) \left(\sec \theta \tan \theta d \theta\right) = \int \theta {\sec}^{3} \theta \tan \theta d \theta$

Note that $\int {\sec}^{3} \theta \tan \theta d \theta = \int {\sec}^{2} \theta \textcolor{w h i t e}{.} d \left(\sec \theta\right) = \frac{1}{3} {\sec}^{3} \theta$. Then, we can easily perform integration by parts on $I$ by letting:

$u = \theta \text{ "=>" } \mathrm{du} = d \theta$

$\mathrm{dv} = {\sec}^{3} \theta \tan \theta d \theta \text{ "=>" } v = \frac{1}{3} {\sec}^{3} \theta$

So:

$I = \frac{1}{3} \theta {\sec}^{3} \theta - \frac{1}{3} \int {\sec}^{3} \theta d \theta \text{ "" "" } \textcolor{red}{\star}$

We can solve this integral with integration by parts again, now letting:

$u = \sec \theta \text{ "=>" } \mathrm{du} = \sec \theta \tan \theta d \theta$

$\mathrm{dv} = {\sec}^{2} \theta d \theta \text{ "=>" } v = \tan \theta$

So:

$I = \frac{1}{3} \theta {\sec}^{3} \theta - \frac{1}{3} \left(\sec \theta \tan \theta - \int \sec \theta {\tan}^{2} \theta d \theta\right)$

Letting ${\tan}^{2} \theta = {\sec}^{2} \theta - 1$ this becomes:

$I = \frac{1}{3} \theta {\sec}^{3} \theta - \frac{1}{3} \sec \theta \tan \theta + \frac{1}{3} \int \sec \theta \left({\sec}^{2} \theta - 1\right)$

$\textcolor{w h i t e}{I} = \frac{1}{3} \theta {\sec}^{3} \theta - \frac{1}{3} \sec \theta \tan \theta + \frac{1}{3} \int {\sec}^{3} \theta d \theta - \frac{1}{3} \int \sec \theta d \theta$

$\textcolor{w h i t e}{I} = \frac{1}{3} \theta {\sec}^{3} \theta - \frac{1}{3} \sec \theta \tan \theta - \frac{1}{3} \ln \left\mid \sec \theta + \tan \theta \right\mid + \frac{1}{3} \int {\sec}^{3} \theta d \theta$

Earlier ($\textcolor{red}{\star}$) we noted that $I = \frac{1}{3} \theta {\sec}^{3} \theta - \frac{1}{3} \int {\sec}^{3} \theta d \theta$, so we can set this to be equivalent to the expression we just found.

$\frac{1}{3} \theta {\sec}^{3} \theta - \frac{1}{3} \int {\sec}^{3} \theta d \theta = \frac{1}{3} \theta {\sec}^{3} \theta - \frac{1}{3} \sec \theta \tan \theta - \frac{1}{3} \ln \left\mid \sec \theta + \tan \theta \right\mid + \frac{1}{3} \int {\sec}^{3} \theta d \theta$

Rearranging:

$- \frac{2}{3} \int {\sec}^{3} \theta d \theta = - \frac{1}{3} \sec \theta \tan \theta - \frac{1}{3} \ln \left\mid \sec \theta + \tan \theta \right\mid$

Which gives:

$- \frac{1}{3} \int {\sec}^{3} \theta d \theta = - \frac{1}{6} \sec \theta \tan \theta - \frac{1}{6} \ln \left\mid \sec \theta + \tan \theta \right\mid$

Plugging this into $\textcolor{red}{\star}$ yields:

$I = \frac{1}{3} \theta {\sec}^{3} \theta - \frac{1}{6} \sec \theta \tan \theta - \frac{1}{6} \ln \left\mid \sec \theta + \tan \theta \right\mid$

We can now return to our original variable $x$ using $x = \sec \theta$, $\tan \theta = \sqrt{{\sec}^{2} \theta - 1} = \sqrt{{x}^{2} - 1}$ and $\theta = {\sec}^{-} 1 \left(x\right)$:

$I = \frac{1}{3} {x}^{3} {\sec}^{-} 1 \left(x\right) - \frac{1}{6} x \sqrt{{x}^{2} - 1} - \frac{1}{6} \ln \left\mid x + \sqrt{{x}^{2} - 1} \right\mid + C$