First slide

Methods of integration

Question

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

Moderate

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

We can write
$\begin{array}{l} I = \int \frac{- 2 \sin^{2} 2 x \cos x \sin x}{\cos^{2} x - \sin^{2} x} d x \\ = - \int \frac{\sin^{3} 2 x}{\cos 2 x} d x = - \int \frac{\sin^{3} 2 x}{1 - \sin^{2} 2 x} \cos 2 x d x \end{array}$
Put $\sin 2 x = t$ to obtain
$\begin{array}{l} I = - \frac{1}{2} \int \frac{t^{3}}{1 - t^{2}} d t = \frac{1}{2} \int \frac{t (1 - t^{2}) - t}{1 - t^{2}} d t \\ = \frac{1}{4} t^{2} + \frac{1}{4} \log |1 - t^{2}| + C \\ = \frac{1}{4} \sin^{2} 2 x + \frac{1}{2} \log | \cos 2 x | + C . \end{array}$

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