# How do you integrate int sec^2sqrtx by integration by parts method?

## How do you integrate $\int {\sec}^{2} \sqrt{x} \mathrm{dx}$ by integration by parts method?

Sep 25, 2016

$2 \left[\sqrt{x} \tan \left(\sqrt{x}\right) - \ln | \tan \sqrt{x} + \sec \sqrt{x} |\right] + C$.

#### Explanation:

Let $\sqrt{x} = y \Rightarrow x = {y}^{2} \Rightarrow \mathrm{dx} = 2 y \mathrm{dy}$.

$\therefore I = \int {\sec}^{2} \sqrt{x} \mathrm{dx} = \int \left({\sec}^{2} y\right) \left(2 y\right) \mathrm{dy} = 2 \int y {\sec}^{2} y \mathrm{dy}$.

The Rule of Integration by Parts (IbP) states :

$\text{ (IbP) : } \int u v \mathrm{dy} = u \int v \mathrm{dy} - \int \left(\frac{\mathrm{du}}{\mathrm{dy}} \cdot \int v \mathrm{dy}\right) \mathrm{dy}$.

We take, $u = y , \mathmr{and} , v = {\sec}^{2} y$.

$\therefore \frac{\mathrm{du}}{\mathrm{dy}} = 1 , \mathmr{and} , \int v \mathrm{dy} = \tan y$.

$\therefore I = 2 \left[y \tan y - \int \tan y \mathrm{dy}\right]$

$= 2 \left[y \tan y - \ln | \tan y + \sec y |\right\}$

Replacing $y \text{ by } \sqrt{x}$, we have,

$I = 2 \left[\sqrt{x} \tan \left(\sqrt{x}\right) - \ln | \tan \sqrt{x} + \sec \sqrt{x} |\right] + C$.

Enjoy maths.!