How do you find the integral of $int cot^4 (3x)dx$ ?

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Answer 1 · 2018-07-19T06:00:00

$intcot^4(3x)dx=1/9[9x+3cot(3x)-cot^3(3x)]+c$

Here ,

$I=intcot^4(3x)dx$

Subst. $color(blue)(3x=u=>3dx=du=>dx=1/3du$

$I=1/3intcot^4udu$

$I=1/3intcot^2u*cot^2udu$

$I=1/3intcot^2u(csc^2u-1)duto[becausecolor(red)(csc^2theta-cot^2theta=1)]$

$I=1/3intcot^2ucsc^2udu-1/3intcot^2udu$

$I=-1/3intcot^2u(-csc^2u)du-1/3intcot^2udu$

$I=-1/3int(cotu)^2d(cotu)-1/3int(csc^2u-1)du$

$I=-1/3(cotu)^3/3-1/3{-cotu-u}+c$

$I=-1/9cot^3u+1/3cotu+1/3u+c$

$I=1/9[3u+3cotu-cot^3u]+c$

Subst. back $color(blue)(u=3x$

$I=1/9[9x+3cot(3x)-cot^3(3x)]+c$

How do you find the integral of int cot^4 (3x)dxint cot^4 (3x)dx?