Results 1 to 4 of 4

Math Help - Integration by substiution?? Don't get??

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    66

    Integration by substiution?? Don't get??

    (secx)^2 sqrt((tanx)^3)dx


    x^3/sqrt(x^2+25)

    (5x^2 + 6x - 12)/{{x^4}}


    Integral from 3 to sqrt(24) of x/sqrt(x^2-8)


    Integral from 1/2 to 5/2 of 1/sqrt(6x+1)


    I don't get any of this.. I have a test of this Tuesday and still have no idea how to do it..
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Reefer View Post
    (secx)^2 sqrt((tanx)^3)dx


    x^3/sqrt(x^2+25)

    (5x^2 + 6x - 12)/{{x^4}}


    Integral from 3 to sqrt(24) of x/sqrt(x^2-8)


    Integral from 1/2 to 5/2 of 1/sqrt(6x+1)


    I don't get any of this.. I have a test of this Tuesday and still have no idea how to do it..
    The concept for most of these is all based on the Chain Rule: \left(g(f(x))\right)'=g'(f(x))\cdot f'(x). So now doesn't it make sense that \int g'(f(x))\cdot f'(x)dx=g(f(x))?

    So lets take your first one

    \int \sec^2(x)\sqrt{\tan^3(x)}dx The key here is to identify your f' . The first thing you should notice is that \left(\tan(x)\right)'=\sec^2(x). So now let us rewrite this as \int\sec^2(x)\sqrt{\tan^3(x)}dx=\int\left(\tan(x)\  right)'\sqrt{\tan^3(x)}dx. The next step is to identify our f since we have that our f'=(\tan(x))' it makes sense that f=\tan(x). So now let us rewrite our integral as \int\left(\tan(x)\right)'\cdot\sqrt{\left\{\tan(x)  \right\}^3}dx. From here we see that g'=\sqrt{x^3} so g=\int\sqrt{x^3}=\frac{2x^{\frac{5}{2}}}{5}. So by the fact that \int g'(f(x))\cdot f'(x)dx=g(f(x)) we can see that \int\sec^2(x)\sqrt{\tan^3(x)}dx=\frac{2\tan^{\frac  {5}{2}}(x)}{5}+C. Now that was a lot of work wasn't it? There is a method that makes this much easier it is called u-substitution. It basically say that if I have \int g'(f(x))\cdot f'(x)dx why not let f(x)=u...but I have only replaced half of my "symbols" in the integral. What about f'(x)dx? We take care of this by noting that if u=f(x)\implies du=f'(x)dx. So after our substitution our integral becomes \int g(u) du which usually makes this much easier. So in the previous example had we let u=\tan(x)\implies du=\sec^2(x)dx our integral would have been transformed into \int\sqrt{u^3}du which is pretty easy. Most of the ones you have here are of the form \int g'(f(x))f'(x)dx...see if you can find them and then see if you can do them. The ones that aren't should be apparent...ask back later for more help on those.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2008
    Posts
    66
    Ok I got how to so the first one but...

    I am still lost on every other one..

    x^3/sqrt(x^2+25)

    I have no idea.. I got my answer as x/sqrt(x^2+25) but the answer is something weird..
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by Reefer View Post
    Ok I got how to so the first one but...

    I am still lost on every other one..

    x^3/sqrt(x^2+25)

    I have no idea.. I got my answer as x/sqrt(x^2+25) but the answer is something weird..
    For \int \frac{x^3}{\sqrt{x^2+25}}, try a sub such as u^2 = x^2+25 \implies 2u~du=2x~dx
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: November 2nd 2010, 04:57 AM
  2. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  3. Replies: 2
    Last Post: February 19th 2010, 10:55 AM
  4. Replies: 6
    Last Post: May 25th 2009, 06:58 AM
  5. rational substiution integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 25th 2008, 12:01 PM

Search Tags


/mathhelpforum @mathhelpforum