Question

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

Answer 1

`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.

Answer 2

`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`