Question

\[\int \tan^3 \text{2x sec 2x dx}\]

Answer 1

`  ∫   tan^3 \text{x 2x . sec  (2x) dx}`
\[ = \int \tan^2 2x . \text{sec 2x  tan 2x dx}\]
`  ∫    ( \sec^2 \left( 2x \right) - 1 \right)   \text{sec (2x) tan (  2x ) dx `
\[\text{Let sec }\left( 2x \right) = t\]
`  ⇒  sec  ( 2x )  tan   (2x)  ×  2 = {dt}/{dx} `
`  ⇒  sec  ( 2x )  tan   (2x) dx = {dt}/{2} `
\[Now, \int \tan^3\text{ x 2x} . \text{sec} \left( \text{2x }\right)dx\]
\[ = \frac{1}{2}\int\left( t^2 - 1 \right) dt\]
\[ = \frac{1}{2}\left[ \frac{t^3}{3} - t \right] + C\]
\[ = \frac{1}{2} \left[ \frac{\sec^3 \left( 2x \right)}{3} - \text{sec}\left( \text{2x }\right) \right] + C\]
\[ = \frac{1}{6} \text{sec}^3 \left( \text{2x} \right) - \frac{\text{sec} \left( \text{2x }\right)}{2} + C\]