Results 1 to 10 of 10

Math Help - Integration by substitution

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    18

    Integration by substitution

    I'm having difficulty with solving this:

    \int_{0}^{\frac{\pi}{2}} \frac{cos\:t}{\sqrt{3+sin\:t}}\: dt

    I'd greatly appreciate any help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008

    Re: Integration by substitution

    Quote Originally Posted by maybealways View Post
    I'm having difficulty with solving this:

    \int_{0}^{\frac{\pi}{2}} \frac{cos\:t}{\sqrt{3+sin\:t}}\: dt

    I'd greatly appreciate any help!
    Make the substitution \displaystyle u = 3 + \sin{t} \implies du = \cos{t}\,dt, and also \displaystyle u(0) = 3 and \displaystyle u\left(\frac{\pi}{2}\right) = 4 and the integral becomes

    \displaystyle \int_3^4{\frac{1}{\sqrt{u}}\,du} = \int_3^4{u^{-\frac{1}{2}}\,du}.

    Can you go from here?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2009
    Posts
    18

    Re: Integration by substitution

    I'm still having difficulty with solving the rest of the equation ^^;;;

    Also, why does \displaystyle u\left(\frac{\pi}{2}\right)?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Integration by substitution

    Like Prove it said the substitution u=3+\sin(t) \Rightarrow du=\cos(t)dt is very useful here.
    Because you have a definite integral you can change the integration limits by using the given substitution.
    The original integration limits are in function of t so the new one has to be in function of u so:
    t=\frac{\pi}{2} \Rightarrow u=3+\sin\left(\frac{\pi}{2}\right)=3+1=4
    t=0 \Rightarrow u=3+\sin(0)=3
    So the new integration limits (in function of u) are 3 and 4.

    To solve the integral use the rule:
    \int_{a}^{b} x^{n} = \left[\frac{x^{n+1}}{n+1}\right]_{a}^{b} and n\neq -1

    Note:
    Changing the integration limits isn't necessary, you can also hold the original integration limits, but afterwards you have to do the back-substitution then.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008

    Re: Integration by substitution

    Quote Originally Posted by maybealways View Post
    I'm still having difficulty with solving the rest of the equation ^^;;;

    Also, why does \displaystyle u\left(\frac{\pi}{2}\right)?
    You are making a substitution \displaystyle u(t) = 3 + \sin{t}, so the notation \displaystyle u(0) means \displaystyle u evaluated at \displaystyle t = 0, and similarly \displaystyle u\left(\frac{\pi}{2}\right) means \displaystyle u evaluated at \displaystyle t = \frac{\pi}{2}. As Siron said, by changing the terminals to \displaystyle u values, you don't need to convert your integral back to a function of \displaystyle t before substituting your terminals.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2009
    Posts
    18

    Re: Integration by substitution

    ^Oh! I see That makes sense (in regards to the the terminal part).

    But, I'm still stuck after everything has been substituted :/ Do I equate both sides separately?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008

    Re: Integration by substitution

    Quote Originally Posted by maybealways View Post
    ^Oh! I see That makes sense (in regards to the the terminal part).

    But, I'm still stuck after everything has been substituted :/ Do I equate both sides separately?
    Equate both sides of what?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Apr 2009
    Posts
    18

    Re: Integration by substitution

    This: \int_3^4{\frac{1}{\sqrt{u}}\,du} = \int_3^4{u^{-\frac{1}{2}}\,du}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008

    Re: Integration by substitution

    Quote Originally Posted by maybealways View Post
    This: \int_3^4{\frac{1}{\sqrt{u}}\,du} = \int_3^4{u^{-\frac{1}{2}}\,du}
    They're two steps of the integration, not an "equation" to solve. Now you need to actually perform the integration using the rule given by Siron in post #4.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Apr 2009
    Posts
    18

    Re: Integration by substitution

    ^Ahh, I misread the question...

    But I'm trouble-free right now! Thanks for the help
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integration by substitution
    Posted in the Calculus Forum
    Replies: 13
    Last Post: December 26th 2010, 10:24 PM
  2. Integration by substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 9th 2009, 05:08 PM
  3. integration by substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 5th 2008, 12:38 PM
  4. Integration by Substitution
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 27th 2008, 05:14 PM
  5. substitution integration
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 6th 2007, 07:51 PM

Search Tags


/mathhelpforum @mathhelpforum