Question:
Evaluate the following integrals:
$\int \frac{2 \cos 2 x+\sec ^{2} x}{\sin 2 x+\tan x-5} d x$
Solution:
Assume $\sin 2 x+\tan x-5=t$
$d(\tan x+\sin 2 x-5)=d t$
$\left(2 \cos 2 x+\sec ^{2} x\right) d x=d t$
Put $t$ and $d t$ in given equation we get
$\Rightarrow \int \frac{d t}{t}$
$=\ln |t|+c$
But $t=\sin 2 x+\tan x-5$
$=\ln |\sin 2 x+\tan x-5|+c$
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