How do you integrate #sin^(3)2xdx#?

2 Answers
Apr 3, 2015

#int sin^3(2x)dx#

I can't integrate straightaway. There's no obvious substitution and no obvious parts.

What do I know about Sine?
#d/(dx)(sinx)=cosx# , #int sinx dx = -cosx#

That's about if for calculus. From trig, there's more:

#sin^2(2x) + cos^2 (2x)x = 1#

so #sin^2 (2x)=1-cos^2(2x)# and I know the derivative of cos is -sin, so maybe a substitution after all. TRY IT.
(I know it will work by many years of experience. A student just has to try something and see if it helps. (So do I, for higher level problems.)

Try it:

#int sin^3(2x)dx = int sin^2 (2x) sin(2x)dx#

#= int (1-cos^2 (2x)) sin (2x) dx = int [sin(2x) - cos^2 (2x) sin(2x)] dx#

Now I should be able to integrate each of these two terms:

#int sin(2x) dx = -1/2 cos (2x)#

and #int - cos^2 (2x) sin(2x) dx#, by letting #u=cos(2x)# we can get

#int - cos^2 (2x) sin(2x) dx = 1/6 cos^3 (2x)#

So we get:

#-1/2 cos (2x)+1/6 cos^3 (2x) +C#

Or see the other solution I'll post.

Apr 3, 2015

#int sin^3 (2x) dx = int (sin(2x))^3 dx = int (2sinx cosx)^3 dx#, So

#int sin^3 (2x) dx = 8int sin^3x cos^3x dx#

Now either pull off #sin^2x# and write the integrand as #(1-cos^2x)cos^3x sinx dx# (So we have powers of sine times differential of sine) or vice versa (details follow)

#8int sin^3x cos^3x dx = 8int sin^3 x (cos^2x) cosx dx# (Do you see where we're going?)

#=8int sin^3 x (1-sin^2x) cosx dx# See it yet?

#=8int (sin^3 x -sin^5x) cosx dx#

#=8int (sin^3 x) cosx dx -int (sin^5x) cosx dx#

#=8(1/4sin^4x - 1/6 sin^6x) +C#

#= 2sin^4x-4/3sin^6x+C#