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Question
\[\int \sec^4 2x \text{ dx }\]
Sum
Solution
\[\int \sec^4 2x \text{ dx }\]
∫ sec4 2x dx
= ∫ sec2 2x . sec2 2x dx
= ∫ (1 + tan2 2x) . sec2 2x dx
Let tan 2x = t
⇒ sec2 2x . 2 dx = dt
\[\Rightarrow \sec^2 2x . dx = \frac{dt}{2}\]
\[Now, \int\left( 1 + \tan^2 2x \right) . \sec^2 2x \text{ dx }\]
\[ = \frac{1}{2}\int\left( 1 + t^2 \right) dt\]
\[ = \frac{1}{2}\left[ t + \frac{t^3}{3} \right] + C\]
\[ = \frac{t}{2} + \frac{t^3}{6} + C\]
\[ = \frac{\tan \left( \text{ 2x } \right)}{2} + \frac{\tan^3 \left( \text{ 2x } \right)}{6} + C\]
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