First slide

Introduction to integration

Question

$Evaluate \int \tan^{4} x d x$

Easy

A

$\frac{\sec^{3} x}{3} - \tan x + C$
B

$\frac{\sec^{3} x}{3} - \tan x + x + C$
C

$\frac{\tan^{3} x}{3} + \tan x + x + C$
D

$\frac{\tan^{3} x}{3} - \tan x + x + C$

Solution

$\begin{array}{l} \int \tan^{4} x d x = \int \tan^{2} x \tan^{2} x d x = \int \tan^{2} x (\sec^{2} x - 1) d x \\ \int \tan^{2} x \sec^{2} x d x - \int \tan^{2} x d x = \frac{\tan^{3} x}{3} - \int (\sec^{2} x - 1) d x \\ = \frac{\tan^{3} x}{3} - \tan x + x + C \end{array}$

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