Question

How do you find the integral of `x^3 * sin( x^2 ) dx`?

Answer 1

`I=1/2(-x^2cosx^2+sinx^2)+C`

Explanation:

`x^2=t => 2xdx=dt, x^3dx=1/2x^2 2xdx = 1/2tdt`

`int x^3 sinx^2 dx = 1/2 int tsintdt = I`

`u=t => du=dt`

`dv=sintdt => v=int sintdt= -cost`

`I=1/2[-tcost + int costdt] = 1/2(-tcost+sint)+C`

`I=1/2(-x^2cosx^2+sinx^2)+C`