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

Answer

Let $\text{I}=\int\sec^4\text{x}\tan\text{x}\text{ dx}\ ....(1)$
Let $\tan\text{x}=\text{t}$ then,
$\Rightarrow\text{d}(\tan\text{x})=\text{dt}$
$\Rightarrow\sec^2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$
Putting $\tan\text{x}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$ in equation (1), we get,
$\text{I}=\int\sec^4\text{x}\tan\text{x}\frac{\text{dt}}{\sec^2\text{x}}$
$=\int\sec^2\text{x}\text{ t dt}$
$=\int\big(1+\tan^2\text{x}\big)\text{t dt}$
$=\int\big(1+\text{t}^2\big)\text{t dt}$
$=\int\big(\text{t}+\text{t}^3\big)\text{dt}$
$=\frac{\text{t}^2}{2}+\frac{\text{t}^4}{4}+\text{C}$
$=\frac{\tan^2\text{x}}{2}+\frac{\tan^4\text{x}}{4}+\text{C}$
$\text{I}=\frac{1}{2}\tan^2\text{x}+\frac{1}{4}\tan^4\text{x}+\text{C}$

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