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Math Help - how can i integrate these using substitution

  1. #1
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    how can i integrate these using substitution

    ive tried everything i could think of and cant get a proper u.


    x^2 (sq.rt.(x+1)) dx

    (sin x) dx/(sq.rt.(cos^3 x))

    sin (sq.rt.(x+1)) dx/ (sq.rt. (x+1) from 3 to 8





    they dont seem like they should cause too much problems, but for the life of me, i cant figure them out

    any help is appreciated, as always
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  2. #2
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    Quote Originally Posted by twostep08 View Post
    ive tried everything i could think of and cant get a proper u.


    x^2 (sq.rt.(x+1)) dx


    By parts! u=x^2\,,\,u'=2x\,,\,\,v'=\sqrt{x+1}\,,\,v=\frac{2}  {3}(x+1)^{3\slash2}\,,\,\,\,so:

    \int x^2\sqrt{x+1}\,dx=\frac{2}{3}x^2(x+1)^{3\slash2}-\frac{4}{3}\int x\sqrt{x+1}\,dx....now you continue (again by parts)!



    (sin x) dx/(sq.rt.(cos^3 x))


    \int\frac{\sin x}{\sqrt{\cos^3x}}dx=\int\sin x\left(\cos^{-3\slash 2}\!(x)\right)dx , and this is an integral of the form \int f'(x)f(x)^n\,dx=\frac{f(x)^{n+1}}{n+1} + C

    OR you can make the substitution u=\cos x\Longrightarrow du=-\sin x \,dx and etc.


    sin (sq.rt.(x+1)) dx/ (sq.rt. (x+1) from 3 to 8


    u=\sqrt{x+1}\Longrightarrow du=\frac{dx}{2\sqrt{x+1}}\,,\,\,and\,\,x=3\Longrig  htarrow u=\sin 2 and etc.

    Tonio



    they dont seem like they should cause too much problems, but for the life of me, i cant figure them out

    any help is appreciated, as always
    .
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  3. #3
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    Quote Originally Posted by twostep08 View Post
    ive tried everything i could think of and cant get a proper u.
    x^2 (sq.rt.(x+1)) dx
    \begin{gathered}<br />
  u = x + 1\; \Rightarrow \;x = u - 1 \hfill \\<br />
  x^2 \sqrt {x + 1} \; \Rightarrow \;\left( {u - 1} \right)^2 \sqrt u  = u^{\frac{5}<br />
{2}}  - 2u^{\frac{3}<br />
{2}}  + u^{\frac{1}<br />
{2}}  \hfill \\ <br />
\end{gathered}
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  4. #4
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    thanks guys...i knew how to do the first by parts, but we had to do it by u-sub...either way, i understand all 3 now--thanks!
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