Question:
Evaluate the following integrals:
$\int \frac{\cos 4 x-\cos 2 x}{\sin 4 x-\sin 2 x} d x$
Solution:
Assume $\sin 4 x-\sin 2 x=t$
$d(\sin 4 x-\sin 2 x)=d t$
$(\cos 4 x-\cos 2 x) d x=d t$
Put $t$ and dt in given equation we get
$\Rightarrow \int \frac{\mathrm{d} t}{t}$
$=\ln |t|+c$
But $t=\sin 4 x-\sin 2 x$
$=\ln |\sin 4 x-\sin 2 x|+c$
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