Find the integral of the function tan3 2x sec 2x

Question

Find the integral of the function tan³ 2x sec 2x

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Answer

Clearly, tan³2x sec 2x = tan²2x tan 2x sec 2x
= (sec²2x - 1) tan 2x sec 2x
= sec²2x.tan 2x sec 2x - tan 2x sec 2x
$\Rightarrow \int \tan ^{3} 2 x \sec 2 x d x=\int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\int \tan 2 x \sec 2 x d x$
$= \int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\frac{\sec 2 x}{2}+C$
Now, Let sec 2x = t
⇒ 2 sec 2x tan 2x dx = dt
Thus, $\int \tan ^{3} 2 x \sec 2 x d x=\frac{1}{2} \int t^{2} d t-\frac{\sec 2 x}{2}+C$
$= \frac{t^{3}}{6}-\frac{\sec 2 x}{2}+C$
$= \frac{(\sec 2 x)^{3}}{6}-\frac{\sec 2 x}{2}+C$

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