Question
Evaluate the following:
$\int\tan^2\text{x}\sec^4\text{x dx}$

Answer

Let $\text{I}=\int\tan^2\text{x}\sec^4\text{xdx}$
$=\int\tan^2\text{x}\sec^2\text{x}\sec^2\text{xdx}$
$=\int\tan^2\text{x}(1+\tan^2\text{x})\sec^2\text{xdx}$
Put $\tan\text{x}=\text{t}\Rightarrow\sec^2\text{x}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\text{t}^2(1+\text{t}^2)\text{dt}$
$=\int(\text{t}^2+\text{t}^4)\text{dt}=\frac{\text{t}^3}{3}+\frac{\text{t}^5}{5}+\text{C}$ $=\frac{\tan^5\text{x}}{5}+\frac{\tan^3\text{x}}{3}+\text{C}$

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