Results 1 to 4 of 4

Math Help - Integral (finished--need someone to check!)

  1. #1
    Newbie
    Joined
    Dec 2010
    Posts
    9

    Integral (finished--need someone to check!)

    the integral is from 0 to π/4:

    (1 + tanθ)^3(sec^2θ)dθ

    So, here's what I did:

    u = 1 + tanθ
    du = sec^2θdθ

    so,

    (u)^3du

    then, I changed the integral limits because I think I'm supposed to (not sure?)

    1 + tanθ(π/4) = 2
    1 + tanθ(0) = 1

    so,

    1/4(u)^4

    then I simply plugged the new limit numbers into the equation to finish it

    [1/4(2)^4] - [1/4(1)^4] = 15/4

    Did I do it right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Quote Originally Posted by OhhCalculus View Post
    the integral is from 0 to π/4:

    (1 + tanθ)^3(sec^2θ)dθ

    So, here's what I did:

    u = 1 + tanθ
    du = sec^2θdθ

    so,

    (u)^3du

    then, I changed the integral limits because I think I'm supposed to (not sure?)

    1 + tanθ(π/4) = 2
    1 + tanθ(0) = 1

    so,

    1/4(u)^4

    then I simply plugged the new limit numbers into the equation to finish it

    [1/4(2)^4] - [1/4(1)^4] = 15/4

    Did I do it right?
    Seems right to me.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Does this give you the same answer?

    \displaystyle \left[ \frac{1}{4}(\tan \theta+1)^4\right]_0^{\frac{\pi}{4}}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,393
    Thanks
    1327
    Quote Originally Posted by OhhCalculus View Post
    the integral is from 0 to π/4:

    (1 + tanθ)^3(sec^2θ)dθ

    So, here's what I did:

    u = 1 + tanθ
    du = sec^2θdθ

    so,

    (u)^3du
    Execellent!

    then, I changed the integral limits because I think I'm supposed to (not sure?)
    You could either
    1) Do the integral in terms of u, then write the answer in terms of \theta again and evaluate between the limits of the original integral.
    2) change the limits of integration to values of u and then evaluate the between those limits.

    In my opinion, it is better and easier to do (2) as you do here.
    Of course, you do NOT use the " \theta" values with the "u" variable!

    1 + tanθ(π/4) = 2
    1 + tanθ(0) = 1

    so,

    1/4(u)^4

    then I simply plugged the new limit numbers into the equation to finish it

    [1/4(2)^4] - [1/4(1)^4] = 15/4

    Did I do it right?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. finished goods
    Posted in the Business Math Forum
    Replies: 1
    Last Post: September 10th 2009, 05:52 PM
  2. int by parts(almost finished)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 22nd 2009, 03:58 PM
  3. Finished - but is it Right (ODE)??
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 27th 2008, 05:42 PM
  4. Have I finished this?
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: May 25th 2008, 03:09 AM
  5. Could somebody check this integral ?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 14th 2007, 04:30 AM

Search Tags


/mathhelpforum @mathhelpforum