First slide

Methods of integration

Question

The integral $\int \frac{\cos 4 x - 1}{\cot x - \tan x}$ is equal to

Easy

A

$- \frac{1}{2} \cos 4 x + C$
B

$- \frac{1}{4} \cos 4 x + C$
C

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

none of these

Solution

Put $\cos 2 x = t$ Then $- 2 \sin 2 x d x = d t$ and the
integral becomes
$\int (\frac{- 2 \sin^{2} 2 x}{\cos^{2} x - \sin^{2} x}) \cos x \sin x d x$
$\begin{array}{l} = - \int (\frac{1 - \cos^{2} 2 x}{\cos 2 x}) \sin 2 x d x \\ = \frac{1}{2} \int \frac{1 - t^{2}}{t} d t = \frac{1}{2} (\log | t | - \frac{t^{2}}{2}) + C \\ = \frac{1}{2} \log | \cos 2 x | - \frac{1}{4} \cos^{2} 2 x + C . \end{array}$

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