Find the integral of the function ^3 2x 2x - Vidyadip

Question

Find the integral of the function $\tan^3 2x \sec 2x$

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Answer

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