Evaluate the following integrals:

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 $t=\sin 2 x+\tan x-5$

$=\ln |\sin 2 x+\tan x-5|+c$