How do you integrate int tan^3(3x)?

Jan 7, 2017

The answer is =1/3ln(∣cos(3x)∣)+1/6sec^2(3x)+C

Explanation:

${\sec}^{2} x = 1 + {\tan}^{2} x$

${\tan}^{2} x = {\sec}^{2} x - 1$

$\tan x = \sin \frac{x}{\cos} x$

$\sec x = \frac{1}{\cos} x$

Let $u = 3 x$, $\implies$, $\mathrm{du} = 3 x \mathrm{dx}$

$\int {\tan}^{3} \left(3 x\right) \mathrm{dx} = \frac{1}{3} \int {\tan}^{3} u \mathrm{du}$

$= \frac{1}{3} \int \left({\sec}^{2} u - 1\right) \tan u \mathrm{du}$

$= \frac{1}{3} \int \left(\frac{1}{\cos} ^ 2 u - 1\right) \cdot \sin \frac{u}{\cos} u \mathrm{du}$

$= \frac{1}{3} \int \frac{1 - {\cos}^{2} u}{{\cos}^{3} u} \sin u \mathrm{du}$

Let $v = \cos u$, $\implies$, $\mathrm{dv} = - \sin u \mathrm{du}$

Therefore,

$= \frac{1}{3} \int \frac{1 - {\cos}^{2} u}{{\cos}^{3} u} \sin u \mathrm{du} = \frac{1}{3} \int - \frac{1 - {v}^{2}}{v} ^ 3 \mathrm{dv}$

$= \frac{1}{3} \int \frac{{v}^{2} - 1}{v} ^ 3 \mathrm{dv}$

$= \frac{1}{3} \int \left(\frac{1}{v} - \frac{1}{v} ^ 3\right) \mathrm{dv}$

$= \frac{1}{3} \left(\ln v + \frac{1}{2 {v}^{2}}\right)$

$= \frac{1}{3} \left(\ln \left(\cos u\right) + \frac{1}{2 {\cos}^{2} u}\right)$

=1/3(ln(∣cos(3x)∣)+1/(2cos^2(3x)))+C

=1/3ln(∣cos(3x)∣)+1/6sec^2(3x)+C