Question

Evaluate the Integral $\int {\mathrm{xsec}}^2\left(\mathrm{x}\right)\mathrm{d}x$

Answer 1

Finding the integral:

Given, $\int {\mathrm{xsec}}^2\left(\mathrm{x}\right)\mathrm{d}x$

Integration by parts is done when conventional methods of integration do not work on the integrand.

The formula for integration by parts is given as,

$\int f\left(x\right)·g\left(x\right)\mathrm{d}x=f\left(x\right)\int g\left(x\right)dx-\int \left(f\text{'}\left(x\right)\int g\left(x\right)\mathrm{d}x\right)\mathrm{d}x$

[General tip: When picking the functions to be substituted into the formula, $f\left(x\right)$ must be a function that is easily differentiable or goes to zero when differentiated repeatedly. Likewise, $g\left(x\right)$ must be the function that is more easily integrated]

Let, $f\left(x\right)=x$ and $g\left(x\right)={\sec }^2\left(x\right)$. Thus,

$\begin{array}{cc}\int x{\sec }^2\left(x\right)\mathrm{d}x=\mathrm{x}\int {\sec }^2\left(x\right)\mathrm{d}x-\int \left[\frac{d}{d\mathrm{x}}x\int {\sec }^2\left(x\right)\mathrm{d}x\right]\mathrm{d}x& \\ =\mathrm{xtan}\mathit{\left(}x\mathit{\right)}-\int \tan \mathit{\left(}x\mathit{\right)}\mathrm{d}x& \left[\because \int {\sec }^2\left(\mathrm{x}\right)\mathrm{d}x=\tan \mathit{\left(}x\mathit{\right)}\right]\\ =\mathrm{xtan}\left(x\right)-\ln \left|\sec \mathit{\left(}x\mathit{\right)}\right|+\mathrm{C}& \left[\because \int \tan \left(x\right)\mathrm{d}x=\ln \left|\sec \left(x\right)\right|\right]\end{array}$

Hence $\int x{\sec }^2\left(x\right)\mathrm{d}x=x\tan \left(x\right)-\ln \left|\sec \left(\mathrm{x}\right)\right|+\mathrm{C}$.