Question

How do you find the integral of `sinpixcospix dx`?

Answer 1

For this integral, you should recall that `\sin(2x)=2\sin(x)\cos(x)`. So, if you multiply the sine and cosine function evaluated in the same point, dividing by two the relation above, you can check that
`{\sin(2x)}/{2}=\sin(x)\cos(x)`.

In your case, instead of `x` you have `\pi x`, so the relations becomes
`\sin(\pi x)\cos(\pi x) = {\sin(2\pi x)}/2`, which is much easier to integrate. In fact,
`\int {\sin(2\pi x)}/2 = 1/2\int \sin(2\pi x)`
let `t` be the new variable:
`t=2\pi x \Rightarrow x=t/{2\pi} \Rightarrow dx = dt/{2\pi}`

The integral becomes
`1/2\int \sin(t)dt/{2\pi}`, and factoring out constants again we get
`1/{4\pi}\int \sin(t)dt=-\cos(t)/{4\pi} +C`
Recalling the relation between `t` and `x`, the answer is finally
`-\cos(2\pi x)/{4\pi} +C`

Answer 2

One method gives correct answer: `-\cos(2\pi x)/{4\pi} +C`

A different method gives an answer that looks different:

`intsin(pix)cos(pix)dx`

Since d(sin t) =cos t dt, substitution will work:

Let `u=sin(pix)`, then `du= pi cos(pix) dx` And `cos(pix)=(du)/(pi)`.

Substituting yields:

`1/(pi)intsinu du=1/(pi)(sin^2u)/2+C`

So
`intsin(pix)cos(pix)dx=1/(2pi)sin^2(pi x)+C`

The answer look different, but what is the difference?
Hint: subtract and simplify.

`1/(2pi)sin^2(pi x)-(-\cos(2\pi x)/{4\pi})`
`=1/(2pi)sin^2(pi x)+1/(4 pi)cos(2 pi x)`
`=1/(2pi)sin^2(pi x)+1/(4 pi)[1-2sin^2( pi x)]`
`=1/(4pi)`

The difference is a constant! The two solutions have different `C`'s.