Question

Can somebody show how to integrate this? x^3(4-x^2)^1/2 dx

Answer 1

do it by parts

∫uv' = uv - ∫u'v

v' = x(4-x^2)^1/2
so u = .....

Answer 2

OMG!! You're so smart! Didnt think of that! Thanks!

Answer 3

it was just a suggestion. i hope it works ;)

Answer 4

You can also use a trigonometric substitution. It's not pretty, but it does the job. Set \(x=2\sin u\), then \(dx=2\cos u\,du\), and
\[\int x^3\sqrt{4-x^2}\,dx=\int 8\sin^3u\sqrt{4-4\sin^2u}\,(2\cos u)\,du=32\int \sin^3u\cos^2u\,du\]
Now, \(\sin^3u\cos^2u=\sin u(\cos^2u-\cos^4u)\) by the Pythagorean identity, so the integral is equivalent to
\[32\int \sin u(\cos^2u-\cos^4u)\,du\]
which simplifies nicely with the substitution \(t=\cos u\) such that \(dt=-\sin u\,du\), giving
\[32\int (t^4-t^2)\,dt\]