How do you integrate #int cos^3(x/3)dx#?

1 Answer
mason m
Jan 9, 2017

#3sin(x/3)-sin^3(x/3)+C#

Explanation:

First let #t=x/3#. This implies that #dt=1/3dx#. We then see that:

#intcos^3(x/3)dx=3intcos^3(x/3)1/3dx=3intcos^3(t)dt#

To do this, split up #cos^3(t)# into #cos^2(t)cos(t)# and then rewrite #cos^2(t)# using the identity #sin^2(theta)+cos^2(theta)=1=>cos^2(theta)=1-sin^2(theta)#.

#3intcos^3(t)dt=3intcos^2(t)cos(t)dt=3int(1-sin^2(t))cos(t)dt#

Now let #s=sin(t)#, so #ds=cos(t)dt#. Luckily we already have this in the integrand!

#3int(1-sin^2(t))cos(t)dt=3int(1-s^2)ds#

Integrating term by term using #ints^nds=s^(n+1)/(n+1)# where #n!=-1#.

#3int(1-s^2)ds=3(s-s^3/3)=3s-s^3#

From #s=sin(t)# and #t=x/3# we see that #s=sin(x/3)#. Also add the constant of integration.

#3s-s^3=3sin(x/3)-sin^3(x/3)+C#

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