Question
Evalute the following integrals:
$\int\frac{2\cos2\text{x}+\sec^2\text{x}}{\sin2\text{x}+\tan\text{x}-5}\text{dx}$

Answer

Let $\text{I}=\int\frac{2\cos2\text{x}+\sec^2\text{x}}{\sin2\text{x}+\tan\text{x}-5}\text{dx}$
Putting $\sin2\text{x}+\tan\text{x}-5=\text{t}$
$\Rightarrow2\cos2\text{x}+\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(2\cos2\text{x}+\sec^2\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}|\sin2\text{x}+\tan\text{x}-5|+\text{C} $ $\big[\because\text{t}=\sin2\text{x}+\tan\text{x}-5\big]$

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