Question

Integration help?

How do I integrate (x^3)(sin(x^2))??

Answer 1

∫ [(x³)(sin(x²))] dx.

Let x² = u

2x dx = du

x dx = du/2.

∫ [(x³)(sin(x²))] dx = ∫ [(x²)(sin(x²))] xdx

= (1/2) ∫ [u sin(u)] du

Use integrate by parts : ∫ f(x)g '(x) dx = f(x)g(x) - ∫ f '(x)g(x) dx

Consider f(x) = u and g(x) = - cos u.

(1/2) ∫ [u sin(u)] du = (1/2)[- ucos u - ∫(- cos u)du]

= (1/2)[- u cos u - (- sin u)] + C

= (1/2)[sin u - u cos u] + C.

Put u = (x²)

= (1/2)[sin(x²)  - x²cos (x²)] + C.

Therefore,

∫ [(x³)(sin(x²))] dx = (1/2)[sin(x²)  - x²cos (x²)] + C.