integrate:
[(1/3)x^3]*(1+x^4)^(1/2)
u = 1 + x^4
du = 4x^3dx
but i have no idea how to finish it from here any help would be greatly appreciated...i took cal 1 about a year ago and i forgot how to properly use u substitution
$\displaystyle \displaystyle \frac{1}{3} \int \frac{x^3}{\sqrt{1+x^4}} \, dx
$
$\displaystyle u = 1 + x^4$
$\displaystyle du = 4x^3 \, dx$
$\displaystyle \displaystyle \frac{1}{12} \int \frac{4x^3}{\sqrt{1+x^4}} \, dx$
$\displaystyle \displaystyle \frac{1}{12} \int \frac{1}{\sqrt{u}} \, du$
can you finish?
$\displaystyle \displaystyle\frac{1}{3}\int{x^3\sqrt{1+x^4}}\;dx$
$\displaystyle \displaystyle\ u=1+x^4\Rightarrow\frac{du}{dx}=4x^3\Rightarrow\fr ac{du}{4}=x^3dx$
$\displaystyle \displaystyle\int{x^3\sqrt{1+x^4}\;dx=\int{\sqrt{1 +x^4}\;x^3dx$
The integral becomes
$\displaystyle \displaystyle\frac{1}{12}\int{u^{\frac{1}{2}}du$
I did ... should be $\displaystyle \displaystyle \frac{1}{3} \int x^3 \sqrt{1+x^4} \, dx$
same sub ...
$\displaystyle \displaystyle \frac{1}{12} \int 4x^3 \sqrt{1+x^4} \, dx$
$\displaystyle \displaystyle \frac{1}{12} \int \sqrt{u} \, du$
thought I saw a negative on the 1/2 exponent.
ive worked my problem down to the last step both of u have...now do i just integrate the simple equation with u? and then plug in for u and du? to get my final awnser?
so for example
if i integrate your last equation i get
(2/3)u^(3/2) du <---- do i just plug in (1+x^4) for u and im done?
Yes, as your final solution needs to be in terms of x.
Also add on the constant of integration.
The u-substitution allows a "convenient" solution.
For the calculation of definate integrals (not necessary here), you may choose to return to x,
or evaluate the definate integral using u with a change of limits to u.
$\displaystyle \displaystyle \begin{aligned} \frac{1}{3}\int{x^3\left(1+x^4}\right)^{\frac{1}{2 }}\;{dx} & = \frac{1}{4(3)}\int{\left(1+x^4}\right)^{\frac{1}{2 }}(1+x^4)'\;{dx} \\& = \frac{2}{3(4)(3)}\int{\left[\left(1+x^4}\right)^{\frac{3}{2}}\right]'\;{dx} \\& = \frac{1}{18}\left(1+x^4}\right)^{\frac{3}{2}}+k.\e nd{aligned}$
@ maybenxtseasn: The Coffee Machine's point relates (I think) to your earlier question about u-substitution and the chain rule... to which the answer was yes, the first one is only for navigating the second. So if the first is bogging down you might want to try a different navigational aid (if only to see how the u-sub should be working). Here's my version of what TCM is pointing out...
To find F, integrate with respect to the dashed balloon (which is like where you integrate with respect to u)...
So you have...
... where (key in spoiler) ...
Spoiler:
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
The purpose of the u-substitution was to make it easy!
$\displaystyle \displaystyle\frac{d}{dx}\left(x^4+C\right)=4x^3\R ightarrow\ \int{d\left(x^4+C\right)}=\int{4x^3}\;dx\Rightarro w\int{4x^3}dx=x^4+C$
$\displaystyle \displaystyle\frac{d}{du}\left(u^4+C\right)=4u^3\R ightarrow\int{4u^3}du=u^4+C$
$\displaystyle \displaystyle\int{\sqrt{1+x^4}\;x^3dx=\int{\sqrt{u }}\;du$
when $\displaystyle u=1+x^4$