Question #a6c6d

2 Answers
May 17, 2017

sin(3x)/3-sin^3(3x)/9+C

Explanation:

First, make a u substitution. Let u=3x. That means du=3dx. We'll have a 1/3 outside the integral:

1/3intcos^3u

Split the cos^3u into cosu and cos^2u:

1/3intcosucos^2u

Use trig identities on cos^2u. cos^2u=1-sin^2u

1/3intcosu(1-sin^2u)

Do another substitution. To not confuse myself, I'll call my variable v. Let v=sinu. That means dv=-cosudu

1/3int1-v^2

This we can integrate. Don't forget the constant:

1/3(v-v^3/3+C)

Distribute the 1/3:

v/3-v^3/9+C

Note that I left C the same even though I distributed the 1/3 into it. Dividing a constant by 3 will still leave a constant.

Undo the substitutions. Remember that v=sinu and u=3x:

sinu/3-sin^3u/9+C

sin(3x)/3-sin^3(3x)/9+C

May 17, 2017

1/36(sin9x+9sin3x)+C.

Explanation:

Recall that, cos3theta=4cos^3theta-3costheta.

Replacing theta by 3x, we have,

cos(3(3x))=4cos^3(3x)-3cos(3x), or,

cos9x+3cos3x=4cos^3(3x).

:. intcos^3(3x)dx=1/4int(cos9x+3cos3x)dx,

=1/4(sin(9x)/9+3sin(3x)/3)

rArr intcos^3(3x)dx=1/36(sin9x+9sin3x)+C.

Enjoy Maths.!