Question
Find $\int\sec^3\text{xdx}.$

Answer

$\int\sec^3\text{xdx}$
$=\int\sec\text{x}\cdot\sec^2\text{xdx}$
$=\int\sqrt{1+\tan^2\text{x}}\cdot\sec^2\text{xdx}$
$\big(\text{Put}\tan\text{x}=\text{t };\sec^2\text{x dx}=\text{dt}\big)$
$=\int\sqrt{1+\text{t}^2}\text{dt}$
$=\frac{\text{t}}{2}\sqrt{1+\text{t}^2}+\frac{1}{2}\log\Big|\text{t}+\sqrt{1+\text{t}^2}\Big|+\text{c}$
$=\frac{\sec\text{x}\cdot\tan\text{x}}{2}+\frac{1}{2}\log\big|\tan\text{x}+\sec\text{x}\big|+\text{c}$

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