Question

How do you integrate `int sin^(4) (2x)`?

Answer 1

`intsin^4(2x)dx=3/8x-1/8sin(4x)+1/64sin(8x)+C`

Explanation:

`I=intsin^4(2x)color(red)(dx)`

A great way to deal with even powers of sine and cosine, assuming a substitution won't work, is to use the cosine double-angle identity.

The form of the cosine double-angle identity involving sine is `cos(2theta)=1-2sin^2(theta)`. This can be solved for `sin^2(theta)` as `sin^2(theta)=(1-cos(2theta))/2`.

Notice that the following are analogous forms:

`sin^2(theta)=(1-cos(2theta))/2" "=>" "color(blue)(sin^2(2x)=(1-cos(4x))/2`

Whatever is in the argument of the cosine function must be double that in the sine function. Using the blue identity, we can then rewrite the integrand:

`I=int(sin^2(2x))^2dx=int((1-cos(4x))/2)^2dx`

From here, expand `(1-cos(4x))^2`:

`I=1/4int(1-2cos(4x)+cos^2(4x))dx`

`I=1/4intdx-1/2intcos(4x)dx+1/4intcos^2(4x)dx`

The first two integrals are solved fairly easily. The second one can be solved using the chain rule in reverse (with the substitution `u=4x=>du=4color(white).dx`).

`I=1/4x-1/8int4cos(4x)dx+1/4intcos^2(4x)dx`

`I=1/4x-1/8sin(4x)+1/4intcos^2(4x)dx`

To resolve this even power of cosine, use another form of the cosine double-angle identity: `cos(2theta)=2cos^2(theta)-1`. This shows that `cos^2(theta)=(1+cos(2theta))/2`.

Which is analogous to:

`cos^2(theta)=(1+cos(2theta))/2" "=>" "color(green)(cos^2(4x)=(1+cos(8x))/2)`

Substituting this in:

`I=1/4x-1/8sin(4x)+1/4int(1+cos(8x))/2dx`

Which can then be split up and integrated as before:

`I=1/4x-1/8sin(4x)+1/8intdx+1/8intcos(8x)dx`

`I=1/4x-1/8sin(4x)+1/8x+1/64int8cos(8x)dx`

`I=3/8x-1/8sin(4x)+1/64sin(8x)+C`