First slide

Methods of integration

Question

If $\begin{array}{l} \int \frac{d x}{\sin 2 x - 2 \sin x} = C - \frac{1}{4} \log | f (x) | + \frac{1}{8 \sin^{2} \frac{x}{2}} \end{array}$ then
$f (x)$ is equal to

Moderate

A

$\tan^{2} x$
B

$2 \tan \frac{x}{2}$
C

$\tan \frac{x}{2}$
D

$\sin \frac{x}{2}$

Solution

$\begin{array}{l} \int \frac{d x}{\sin 2 x - 2 \sin x} = \frac{1}{2} \int \frac{d x}{\sin x (\cos x - 1)} \\ = \frac{1}{2} \int \frac{\sin x d x}{(1 - \cos^{2} x) (\cos x - 1)} \\ = \frac{1}{2} \int \frac{- \sin x d x}{(1 - \cos x)^{2} (1 + \cos x)} \end{array}$
$\begin{array}{l} = \frac{1}{2} \int \frac{d t}{(1 - t)^{2} (1 + t)} (t = \cos x) \\ = \frac{1}{4} \int (\frac{1}{2} \frac{1}{1 + t} + \frac{1}{2} \frac{1}{1 - t} + \frac{1}{(1 - t)^{2}}) d t \end{array}$
$\begin{array}{l} = \frac{1}{4} [\frac{1}{2} \log | 1 + t | - \frac{1}{2} \log | 1 - t | + \frac{1}{1 - t}] + C \\ = \frac{1}{4} [\frac{1}{1 - \cos x} + \frac{1}{2} \log |\frac{1 + \cos x}{1 - \cos x}|] + C \\ = \frac{1}{8} \frac{1}{\sin^{2} \frac{x}{2}} - \frac{1}{8} \log |\frac{2 \sin^{2} x / 2}{2 \cos^{2} x / 2}| + C \\ = \frac{1}{8} \frac{1}{\sin^{2} x / 2} - \frac{1}{4} \log | \tan x / 2 | + C \end{array}$

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