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: Integrate exactly

  1. Jul 21, 2007 #1
    URGENT: Integrate exactly

    1. The problem statement, all variables and given/known data
    Using the substitution [tex]x = \sec ^2 u[/tex], evaluate [tex]\int_2^4 {\frac{{dx}}{{x^2 \sqrt {x - 1} }}} [/tex] exactly

    2. The attempt at a solution
    I have used the substitution and ended up with (after some simplification)
    [tex]\int_2^4 { - 2\cos ^2 u.\sin u.\sqrt {\sec ^2 u - 1} }[/tex]. What do i do now? How do I integrate that, I am very confused on how to go about this now. There are others which I need to solve, but are harder than this one, so I will post them in the future once i understand this one and if i still need help with them. Many thanks to those who help
    unique_pavadrin
     
  2. jcsd
  3. Jul 21, 2007 #2
    [tex]sec^2u-1=tan^2u[/tex]. That gives you [tex]-2cosu sin^2u[/tex]. Substitute t=sinu, dt=cosu du[/tex]. You should be able to do the rest.
     
  4. Jul 21, 2007 #3

    VietDao29

    User Avatar
    Homework Helper

    Well, when you perform a u-substitution, you should also change the lower limit, and upper limit for the integral.

    x = sec2u, where u is restricted to be on the interval [tex]\left[ 0 ;\ \frac{\pi}{2} \right[[/tex], so that tan(u) can be positive.

    x = sec2u ~~~> dx = (2 sinu)/(cos3u) du
    x = 2 ~~~> 1/cos2u = 2 ~~~> cos2(u) = 1/2 ~~~> cos(u) = 1/sqrt(2) ~~~> u = pi/4

    x = 4 ~~~> 1/cos2u = 4 ~~~> cos2(u) = 1/4 ~~~> cos(u) = 1/2 ~~~> u = pi/3

    So, your integral will become:

    [tex]\int_{\frac{\pi}{4}} ^ \frac{\pi}{3} \ \cos ^ 4 u \frac{2 \sin u}{\cos ^ 3 u} \frac{du}{\sqrt{\sec ^ 2 u - 1}} = 2 \int_{\frac{\pi}{4}} ^ \frac{\pi}{3} \ \cos u \sin u \frac{du}{\sqrt{\tan ^ 2 u}}[/tex]

    [tex]= 2 \int_{\frac{\pi}{4}} ^ \frac{\pi}{3} \ \cos u \sin u \frac{du}{|\tan u|}[/tex]

    Since [tex]u \in \left[ 0 ;\ \frac{\pi}{2} \right[[/tex], tan(u) will be non-negative, so, we have:

    [tex]= 2 \int_{\frac{\pi}{4}} ^ \frac{\pi}{3} \ \cos u \sin u \frac{du}{\tan u} = ...[/tex]

    Can you go from here? :)
     
  5. Jul 21, 2007 #4
    great help people, i understand now yay! :simile: thanks a lot
     
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