Question

What is `int -cosx^2/x`?

Answer 1

`-1/2"Ci"(x^2)+C`

Explanation:

We will use `"Ci"(x)`, the special cosine integral function.

`int-cos(x^2)/xdx`

Let `t=x^2` so `dt=2xdx`:

`=-1/2int(cos(x^2))/x^2(2xdx)=-1/2intcos(t)/tdt`

Which is the cosine integral:

`=-1/2"Ci"(t)=-1/2"Ci"(x^2)+C`