Question

How do you find the integral of x^3*sinx^2 dx?

Answer 1

u= x^2 du = 2xdx so (1/2)du= xdx.

and also break up your integral as x*x^2sin(x^2)dx so your substitution would look like

(1/2)usin(u)du and now apply integration by parts

Answer 2

Thank you so much! :)