Integration problems giving me a hard time

[roostercogburn]

13) The integral of x/[ x^2 + x^(2/3)]

This one here I tried letting x= u^3 which I simplified to
int 3u^3/[u^3 - 1] dx than I tried long division, after by parts and then I gave up. Probably whole approach is wrong.

9) int csc^3(x) dx
This one I tried using identities and then I tried the uv - int vdu but am not quite sure how to start this one.

Even if you can tell me the method I need to use that'd be nice.[/code]

 

[ChaoticLlama]

csc³x is a nasty one to integrate.

begin with Integration by parts

\(\displaystyle \L\int {\csc ^3 xdx}\)

let
u = cscx
du = -cscx*cotxdx


dv = csc²xdx
v = -cotx

\(\displaystyle \L\begin{array}{l}
\int {\csc ^3 xdx} = - \cot x\csc x - \int {\csc x\cot ^2 xdx} \\
\int {\csc ^3 xdx} = - \cot x\csc x - \int {(\csc ^3 x - \csc x)dx} \\
\int {\csc ^3 xdx} = - \cot x\csc x - \int {\csc ^3 xdx + \int {\csc xdx} } \\
\end{array}\)

you should be able to take it from here.

 

[soroban]

Hello, roostercogburn!

\(\displaystyle 13)\;\;\L\int\frac{x\,dx}{x^2\,+\,x^{\frac{2}{3}}}\)

Your approach was correct, but long division is not required . . .

\(\displaystyle \text{Let }\,u \,=\,x^{\frac{1}{3}}\;\;\Rightarrow\;\;x\,=\,u^3\;\;\Rightarrow\;\;dx\,=\,3u^2\,du\)

\(\displaystyle \text{Substitute: }\L\,\int\frac{u^3(3u^2\,du)}{u^6\,+\,u^2}\;=\;3\int\frac{u^5\,du}{u^2(u^4\,+\,1)} \;= \;3\int\frac{u^3}{u^4\,+\,1}\,du\)

\(\displaystyle \text{Now let }\,v\,=\,u^4\,+\,1\)

 

[roostercogburn]

Thanks. I came on here and then finished integrating and then differentiated and I got the right answer. yeeeeees.