Question

Evaluate :
$\int ta{n}^{3}2xsec2xdx$.

Answer 1

$\int {\tan }^{3}2x{\sec }^{2}xdx$

$=\int {\left(\frac{\sin 2x}{\cos 2x}\right)}^3\frac{1}{\cos 2x}dx$ $\left[\begin{array}{c}\because \tan \theta =\frac{\sin \theta }{\cos \theta }\\ \because \sec \theta =\frac{1}{\cos \theta }\end{array}\right]$

$=\int \frac{{\sin }^{3}2x}{{\cos }^{4}2x}dx$

$=\int \frac{{\sin }^{2}2x\sin 2x}{{\cos }^{4}2xdx}dx$

$=\int \frac{(1-{\cos }^{2}2x)}{{\cos }^{4}2x}\sin 2xdx$ ……$\left(1\right)$

Let $\cos 2x=t$

on differentiating, we get

$-2\sin 2xdx=dt$

$\therefore \sin 2xdx=-dt/2$

Substituting the above in equation $\left(1\right)$

$=\int \frac{(1-t^2)}{t^4}(-\frac{dt}{2})$

$=\int \left(\frac{t^2-1}{t^4}\right)\frac{dt}{2}$

$=\frac{1}{2}\int (t^{-2}-t^{-4})dt$

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

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

Re-substituting $t=\cos 2x$ in above equation

$=\frac{1}{2}\frac{1}{6{\cos }^{3}2x}-\frac{1}{2\cos 2x}+c$

$=\frac{1}{6{\cos }^{3}2x}-\frac{1}{2\cos 2x}+c$.