First slide

Introduction to integration

Question

$\int \frac{\sin^{4} x}{\cos^{2} x} d x is equal to$

Moderate

A

$- \frac{3 x}{2} + \frac{1}{4} \sin 2 x - 2 \tan x + C$
B

$- \frac{3 x}{2} - \frac{1}{4} \sin 2 x + \tan x + C$
C

$- \frac{3 x}{2} + \frac{1}{4} \sin 2 x + \tan x + C$
D

$- \frac{3 x}{2} - \frac{1}{4} \sin 2 x - 2 \tan x + C$

Solution

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

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