First slide

Methods of integration

Question

$Evaluate \int \frac{1 - \cos x}{\cos x (1 + \cos x)} d x$

Difficult

A

$\log | \sec x + \tan x | - 2 \tan x / 2 + C$
B

$\log | \sec x - \tan x | - 2 \tan x / 2 + C$
C

$\sec x \tan x - 2 \tan x / 2 + C$
D

None of these

Solution

$Let I = \int \frac{1 - \cos x}{\cos x (1 + \cos x)} d x$
$L e t c o s x = y, t h e n \begin{array}{l} \frac{1 - \cos x}{\cos x (1 + \cos x)} = \frac{1 - y}{y (1 + y)} = \frac{1}{y} - \frac{2}{1 + y} \end{array}$
$\begin{array}{l} = \frac{1}{\cos x} - \frac{2}{1 + \cos x} \\ \begin{array}{l} ∴ l = \int \frac{1 - \cos x}{\cos x (1 + \cos x)} d x \\ = \int \frac{1}{\cos x} d x - \int \frac{2}{1 + \cos x} d x \\ = \int \sec x d x - \int \frac{2}{2 \cos^{2} \frac{x}{2}} d x \\ = \int \sec x - \int \sec^{2} \frac{x}{2} d x \\ = \log | \sec x + \tan x | - 2 \tan \frac{x}{2} + C \end{array} \end{array}$

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