Question

Need help evaluating this following integral

Answer 1

\[\int\limits_{}^{} x \sin^3(x^2)dx\] any ideas?

Answer 2

Use substitution with u = x^2, 1/2 du = x

Answer 3

Honestly not sure how that would help though. How would I get rid of the sin^3?

Answer 4

Reduction formula

Answer 5

Never heard of a reduction formula...

Answer 6

As Guzman said, use a u-sub and you get

u = x^2
du = 2xdx

\[1/2 \int\limits_{?}^{?} \sin^3(u) du\]

Don't you agree we can just break this up into:

\[1/2 \int\limits_{?}^{?} \sin^2(u) \sin(u) du\]
by the pythagorean identity Sin^2 (u) = (1 - Cos^2(u))

so we just use that and we get:

\[1/2 \int\limits_{?}^{?} (1 - \cos^2(u)) \sin(u) du\]

Now another u-substitution, lets let:

w = cos(u)

dw = -sin(u) du

so now our integral is just

\[-1/2 \int\limits_{?}^{?} (1-w^2) dw\]

integrating that, we get:

\[(-1/2) (w - (1/3)w^3)\]

back substituting we get:

\[-1/2 (\cos(u) - (1/3)\cos^3(u))\]

= \[-1/2 (\cos(x^2) - (1/3)\cos^3(x^2))\]

happy integrating ;)

Answer 7

Thank you very much! I can see it now :) I was just struggling to get rid of the x infront