How do you integrate $\int {tan}^{3} (3 x)$ $int tan^3(3x)$ ?

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How do you integrate $\int {tan}^{3} (3 x)$ $int tan^3(3x)$ ?

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Answer 1 · 2017-01-07T08:10:22

The answer is $= \frac{1}{3} ln (∣ cos (3 x) ∣) + \frac{1}{6} {sec}^{2} (3 x) + C$ $=1/3ln(∣cos(3x)∣)+1/6sec^2(3x)+C$

Explanation:

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

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

$tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$

$sec x = \frac{1}{cos x}$ $secx=1/cosx$

Let $u = 3 x$ $u=3x$ , $\Rightarrow$ $=>$ , $d u = 3 x d x$ $du=3xdx$

$\int {tan}^{3} (3 x) d x = \frac{1}{3} \int {tan}^{3} u d u$ $inttan^3(3x)dx=1/3inttan^3udu$

$= \frac{1}{3} \int ({sec}^{2} u - 1) tan u d u$ $=1/3int(sec^2u-1)tanudu$

$= \frac{1}{3} \int (\frac{1}{{cos}^{2} u} - 1) \cdot \frac{sin u}{cos u} d u$ $=1/3int(1/cos^2u-1)*sinu/cosudu$

$= \frac{1}{3} \int \frac{1 - {cos}^{2} u}{{cos}^{3} u} sin u d u$ $=1/3int(1-cos^2u)/(cos^3u)sinudu$

Let $v = cos u$ $v=cosu$ , $\Rightarrow$ $=>$ , $d v = - sin u d u$ $dv=-sinudu$

Therefore,

$= \frac{1}{3} \int \frac{1 - {cos}^{2} u}{{cos}^{3} u} sin u d u = \frac{1}{3} \int - \frac{1 - v^{2}}{v^{3}} d v$ $=1/3int(1-cos^2u)/(cos^3u)sinudu=1/3int-(1-v^2)/v^3dv$

$= \frac{1}{3} \int \frac{v^{2} - 1}{v^{3}} d v$ $=1/3int(v^2-1)/v^3dv$

$= \frac{1}{3} \int (\frac{1}{v} - \frac{1}{v^{3}}) d v$ $=1/3int(1/v-1/v^3)dv$

$= \frac{1}{3} (ln v + \frac{1}{2 v^{2}})$ $=1/3(lnv+1/(2v^2))$

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

$= \frac{1}{3} (ln (∣ cos (3 x) ∣) + \frac{1}{2 {cos}^{2} (3 x)}) + C$ $=1/3(ln(∣cos(3x)∣)+1/(2cos^2(3x)))+C$

$= \frac{1}{3} ln (∣ cos (3 x) ∣) + \frac{1}{6} {sec}^{2} (3 x) + C$ $=1/3ln(∣cos(3x)∣)+1/6sec^2(3x)+C$

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