Question

How do I find the integral of x^5(sqrt(4-x^3))

Answer 1

This?
\(\Large \int x^5\sqrt{4-x^3}~dx\)?

Answer 2

Yes! thanks

Answer 3

What have you tried so far?

Answer 4

\[du = \frac{ 3x }{ 2\sqrt{4-x ^{3}} } dx\]

\[du =-3x ^{2}dx\]

\[du =5x ^{4}dx\]

Answer 5

@amistre64 could help you :3

Answer 6

gonna have to do u sub and by parts together

Answer 7

let u = 4-x^3

du = -3x^2 dx

-du/3 = x^2 dx

\[ \int x^3\sqrt{u}~du\]

well, x^3 = 4-u

\[ \int (4-u)\sqrt{u}~du\]
by parts wont be needed, just a few powers is all

Answer 8

forgot the -1/3 .... but thats the basic process id do

Answer 9

How did you turn x^5 to 4-u when x^3=4-u

Answer 10

amistre let
\[ u = 4-x^3 \\
du = -3x^2 dx \]
so
\[ dx = -\frac{1}{3} \frac{du}{x^2} \]
and
\[ \int x^5 \sqrt{4-x^3} \ dx\rightarrow \int x^5 \sqrt{u}\left(-\frac{1}{3}\right) \frac{du}{x^2} \]
which simplifies to
\[ -\frac{1}{3} \int \frac{x^5}{x^2} \sqrt{u} \ du \\ -\frac{1}{3} \int x^3 \sqrt{u} \ du \]

using
\[ u = 4-x^3 \]
and solving for x^3 we get
\[ x^3 = 4-u \]
and fnally
\[ -\frac{1}{3} \int (4-u) \sqrt{u} \ du \\
-\frac{1}{3} \int 4u^\frac{1}{2}-u^\frac{3}{2} \ du
\]

Answer 11

Oh wow ofcourse. Thanks a lot all. I really appreciate this effort!

Answer 12

:)