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

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Answer 1 · 2017-01-07T08:10:22

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

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

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

$tanx=sinx/cosx$

$secx=1/cosx$

Let $u=3x$ , $=>$ , $du=3xdx$

$inttan^3(3x)dx=1/3inttan^3udu$

$=1/3int(sec^2u-1)tanudu$

$=1/3int(1/cos^2u-1)*sinu/cosudu$

$=1/3int(1-cos^2u)/(cos^3u)sinudu$

Let $v=cosu$ , $=>$ , $dv=-sinudu$

Therefore,

$=1/3int(1-cos^2u)/(cos^3u)sinudu=1/3int-(1-v^2)/v^3dv$

$=1/3int(v^2-1)/v^3dv$

$=1/3int(1/v-1/v^3)dv$

$=1/3(lnv+1/(2v^2))$

$=1/3(ln(cosu)+1/(2cos^2u))$

$=1/3(ln(∣cos(3x)∣)+1/(2cos^2(3x)))+C$

$=1/3ln(∣cos(3x)∣)+1/6sec^2(3x)+C$

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