Question

\[\int \sec^4 2x \text{ dx }\]

Answer 1

\[\int \sec^4 2x \text{ dx }\]

∫ sec⁴ 2x dx
=​ ∫ sec² 2x . sec² 2x dx
= ∫ (1 + tan² 2x) . sec² 2x  dx 

Let tan 2x = t
⇒ sec² 2x . 2 dx = dt

\[\Rightarrow \sec^2 2x . dx = \frac{dt}{2}\]
\[Now, \int\left( 1 + \tan^2 2x \right) . \sec^2 2x \text{ dx }\]
\[ = \frac{1}{2}\int\left( 1 + t^2 \right) dt\]
\[ = \frac{1}{2}\left[ t + \frac{t^3}{3} \right] + C\]
\[ = \frac{t}{2} + \frac{t^3}{6} + C\]
\[ = \frac{\tan \left( \text{ 2x } \right)}{2} + \frac{\tan^3 \left( \text{ 2x } \right)}{6} + C\]