First slide

Introduction to integration

Question

$\int \frac{1}{\cos^{2} x \sin^{4} x} d x =$

Moderate

A

$\cot x - 2 \tan x - \frac{\cot^{3} x}{3} + C$
B

$\cot x + 2 \tan x - \frac{\cot^{3} x}{3} + c$
C

$\cot x + 2 \tan x + \frac{\tan^{3} x}{3} + c$
D

$\tan x - 2 \cot x - \frac{\cot^{3} x}{3} + c$

Solution

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

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