Question

Compute ∫1 -1 (i don't know how to write this properly in open study) 2x cos(x2)dx

Answer 1

\[\int 1-2x\cos(x^2)\,dx\] is this your question?

Answer 2

if so,
use the substitution\[u=x^2\]
then,
\[du=2xdx\]
now back to the integral

\[\int 1-2x\cos(x^2)\,dx=\int\,dx-\int \cos(u)\,du=x-\sin(u)+c=x-\sin(x^2)+c \]