# Question #61aa2

Nov 18, 2016

$\int {\tan}^{3} x {\sec}^{2} x \mathrm{dx} = \frac{1}{4} {\tan}^{4} x + C$

#### Explanation:

$\int {\tan}^{3} x {\sec}^{2} x \mathrm{dx}$

the key here is to remember that

$\frac{d}{\mathrm{dx}} \left({\tan}^{n} x\right) = n {\tan}^{n - 1} x {\sec}^{2} x$

**see below**

so comparing this with the required integral

try $\frac{d}{\mathrm{dx}} \left({\tan}^{4} x\right)$

$= 4 {\tan}^{3} x {\sec}^{2} x$

$\int {\tan}^{3} x {\sec}^{2} x \mathrm{dx} = \frac{1}{4} {\tan}^{4} x + C$

proof

$\frac{d}{\mathrm{dx}} \left({\tan}^{n} x\right) = n {\tan}^{n - 1} x {\sec}^{2} x$

$y = {\tan}^{n} x$

$u = \tan x \implies \frac{\mathrm{du}}{\mathrm{dx}} = {\sec}^{2} x$

$y = {u}^{n} \implies \frac{\mathrm{dy}}{\mathrm{du}} = n \left({u}^{n - 1}\right)$

by the chain rule

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = n \left({u}^{n - 1}\right) \times {\sec}^{2} x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = n {\tan}^{n - 1} x \times {\sec}^{2} x$

$= n {\tan}^{n - 1} x {\sec}^{2} x$