Question

How do you find the integral of `x(sin^2(ax))`?

Answer 1

`(2a^2x^2-2axsin(2ax)-cos(2ax))/(8a^2)+C`

Explanation:

`I=intxsin^2(ax)dx`

First, let `t=ax`. This implies that `dt=adx`. Note that this means that `x=t/a` and `dx=(dt)/a`.

`I=intt/asin^2(t)dt/a=1/a^2inttsin^2(t)dt`

This will become easier if we rewrite `sin^2(t)` to make it easier to manipulate, a common trick in integrals of this sort. We will use one version of the cosine double angle formula:

`cos(2t)=1-2sin^2(t)" "=>" "sin^2(t)=1/2(1-cos(2t))`

Using this equivalency, the integral becomes:

`I=1/a^2intt/2(1-cos(2t))dt`

`I=1/(2a^2)int(t-tcos(2t))dt`

Splitting up the integral:

`I=1/(2a^2)inttdt-1/(2a^2)inttcos(2t)dt`

The first integral can be performed easily using the power rule for integration:

`I=1/(2a^2)(t^2/2)-1/(2a^2)inttcos(2t)dt`

`I=t^2/(4a^2)-1/(2a^2)inttcos(2t)dt`

This is another technically unnecessary substitution, but it clears up the following steps. Let `s=2t`, implying that `ds=2dt`. Thus, `dt=(ds)/2` and `t=s/2`.

`I=t^2/(4a^2)-1/(2a^2)ints/2cos(s)(ds)/2`

`I=t^2/(4a^2)-1/(8a^2)intscos(s)ds`

For this integral, we will use integration by parts, which takes the form `intudv=uv-intvdu`. We will let:

`{(u=s" "=>" "du=ds),(dv=cos(s)ds" "=>" "v=sin(s)):}`

Plugging these into the formula:

`I=t^2/(4a^2)-1/(8a^2)[ssin(s)-intsin(s)ds]`

Since `intsin(s)ds=-cos(s)`:

`I=t^2/(4a^2)-1/(8a^2)(ssin(s)+cos(s))`

`I=t^2/(4a^2)-(ssin(s))/(8a^2)-cos(s)/(8a^2)`

Back-substituting with `s=2t`:

`I=t^2/(4a^2)-(2tsin(2t))/(8a^2)-cos(2t)/(8a^2)`

`I=t^2/(4a^2)-(tsin(2t))/(4a^2)-cos(2t)/(8a^2)`

Back-substituting with `t=ax`:

`I=(ax)^2/(4a^2)-(axsin(2ax))/(4a^2)-cos(2ax)/(8a^2)`

`I=(a^2x^2)/(4a^2)-(xsin(2ax))/(4a)-cos(2ax)/(8a^2)`

`I=(x^2)/4-(xsin(2ax))/(4a)-cos(2ax)/(8a^2)`

If we want a common denominator:

`I=(2a^2x^2-2axsin(2ax)-cos(2ax))/(8a^2)+C`