Browse Questions

# Find the integrals of the functions$\tan^32x\sec2x$

Toolbox:
• $\tan^2x=\sec^2x-1.$
• $(ii)\int \sec^2xdx=\tan x+c.$
Given $I=\int\tan^32x\sec2xdx.$

$\tan^32x$ can be written as $\tan^22x\tan 2x$.

$\;\;\;=\int\tan^22x.\tan2x.\sec 2xdx.$

But $\tan^22x=\sec^22x-1.$

$\;\;\;=\int[(\sec^22x-).\tan2x.\sec 2x]dx.$

On multiplying,

$\;\;\;=\int\sec^22x.\tan2x.\sec 2x-\int\sec2x.\tan 2xdx.$

Put $\sec2x=t.$

On differentiating

$2\sec2x\tan 2xdx=dt.$

$(\sec2x\tan 2x)dx=\frac{dt}{2}.$

On substituting we get,

$I=\frac{1}{2}\int t^2 dt-\int\sec2x\tan 2xdx.$

On integrating we get,

$\;\;\;=\frac{1}{2}[\frac{t^3}{3}]-\frac{\sec 2x}{2}+c.$

Substituting for t we get,

$\;\;\;=-\frac{1}{2}(\frac{\sec2x}{3})^3-\frac{\sec2x}{2}+c.$

$\;\;\;=\frac{(\sec2x)^3}{6}-\frac{\sec2x}{2}+c.$