Evaluate the following integrals: ^42x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\sec^42\text{x}\text{ dx}$

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Answer

$\int\sec^42\text{x}\text{ dx}$
$=\int\sec^22\text{x}.\sec^22\text{x}\text{ dx}$
$=\int(1+\tan^22\text{x}).\sec^22\text{x}\text{ dx}$
Let $\tan2\text{x}=\text{t}$
$\sec^22\text{x}.2\text{dx}=\text{dt}$
$\sec^22\text{x}.\text{dx}=\frac{\text{dt}}{2}$
Now, $\int(1+\tan^22\text{x}).\sec^22\text{x}\text{ dx}$
$=\frac{1}{2}\int(1+\text{t}^2)\text{dt}$
$=\frac{1}{2}\Big[\text{t}+\frac{\text{t}^3}{3}\Big]+\text{C}$
$=\frac{\text{t}}{2}+\frac{\text{t}^3}{6}+\text{C}$
$=\frac{\tan(2\text{x})}{2}+\frac{\tan^3(2\text{x})}{6}+\text{C}$

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