Question

What is `int_-pi^pixsinxdx`?

Answer 1

2.

Explanation:

Write `f : x mapsto x sin(x)`.

Because `f(-x) = f(x)` for all `x in [-pi , pi]`, you can say that
`int_(-pi)^pi f(x) dx = 2 int_0^pi f(x) d x`.

Now, use integration by parts with
`u (x) = x` and `v(x) = -cos(x)`.
The functions `u,v` are `C^1` on `[0,pi]` and
`u'(x) = 1` and `v'(x) = sin(x)`.

The integration by parts says that
`int_0^pi u(x)v'(x) dx = [u(x)v(x)]_0^pi - int_0^pi u'(x)v(x) dx`,
therefore,
`int_0^pi x sin(x) dx = [-x cos(x)]_0^pi + int_0^pi cos(x) dx`
`int_0^pi x sin(x) dx = 1 + [sin(x)]_0^pi = 1`.

Finally, `int_(-pi)^pi x sin(x) dx =2int_0^pi x sin(x) dx = 2`.