Question
Evaluate the following integrals:
$\int\tan^5\text{x}\text{ dx}$

Answer

Let $\text{I}=\int\tan^5\text{x}\text{ dx}$ Then
$\text{I}=\int\tan^2\text{x}\tan^3\text{x}\text{ dx}$
$=\int(\sec^2\text{x}-1)\tan^3\text{x}\text{ dx}$
$=\int\sec^2\text{x}\tan^3\text{x}\text{ dx}-\int\tan^3\text{x}\text{ dx}$
$=\int\sec^2\tan^3\text{x}\text{ dx}-\int(\sec^2\text{x}-1)\tan\text{x}\text{ dx}$
Substituting $\tan\text{x}=\text{t}$ and $\sec^2\text{x}\text{ dx}=\text{dt}$ in first two integral, we get
$\text{I}=\int\text{t}^3\text{dt}-\int\text{tdt}+\int\tan\text{x}\text{ dx}$
$=\frac{\text{t}^4}{4}-\frac{\text{t}^2}{2}+\log|\sec\text{x}|+\text{C}$
$=\frac{\tan^4\text{x}}{4}-\frac{\tan^2\text{x}}{2}+\log|\sec\text{x}|+\text{C}$
$\therefore\ \text{I}=\frac{\tan^4\text{x}}{4}-\frac{\tan^2\text{x}}{2}+\log|\sec\text{x}|+\text{C}$

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