Results 1 to 7 of 7

Math Help - Integration via substitution

  1. #1
    Member
    Joined
    Sep 2007
    Posts
    81

    Integration via substitution

    Hi all,

    I've got a bit of a problem with this question, if anyone could show the working it would be appreciated. I've got the suspicion it's meant to be done via substitution, but looking to learn.

    \int^{\pi/8}_0 (\theta + \cos4\theta) d\theta

    I'm pretty sure that the answer is (32 + pi^2) / 128 but of course, the answer isn't everything.

    Anyone able to help me out? Cheers.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2009
    Posts
    80
    Quote Originally Posted by Peleus View Post
    Hi all,

    I've got a bit of a problem with this question, if anyone could show the working it would be appreciated. I've got the suspicion it's meant to be done via substitution, but looking to learn.

    \int^{\pi/8}_0 (\theta + \cos4\theta) d\theta

    I'm pretty sure that the answer is (32 + pi^2) / 128 but of course, the answer isn't everything.

    Anyone able to help me out? Cheers.
    Break it up into the sum of two integrals. One can be done directly, the other via a simple substitution.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2007
    Posts
    81
    Here's my working so far, however I think I've done something wrong.

    <br />
\int^{\pi/8}_0 (\theta + \cos4\theta) d\theta<br />

    <br />
\int^{\pi/8}_0 \theta d\theta + \int^{\pi/8}_0 \cos4\theta d\theta<br />


    <br />
\int^{\pi/8}_0 \theta d\theta + \cos4\int^{\pi/8}_0 \theta d\theta<br />

    <br />
[\frac{\theta^2}{2}]^{\pi/8}_0 + [\cos2\theta^2]^{\pi/8}_0<br />
    Last edited by mr fantastic; October 19th 2009 at 12:36 AM. Reason: Restored original post
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2009
    Posts
    80
    It's cos(4theta), presumably, not cos(4)*theta. You can't pull it out of the integral.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Peleus View Post
    [snip]
    <br />
\int^{\pi/8}_0 \theta d\theta + \cos4\int^{\pi/8}_0 \theta d\theta<br />

    [snip]
    And would you also say that \int \sqrt{x} \, dx = \sqrt{{\color{white} ..}} \int x \, dx. I sincerely hope not! You need to thoroughly review functions as a matter of urgency.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Sep 2007
    Posts
    81
    Yeah looking back it's quite silly of me. Brain snap.

    This could be just as silly, but I think it's better.

    Let 4\theta = x and  dx = 4d\theta

    <br />
\int^{\pi/8}_0 \cos x dx<br />

    <br />
 \int^{\pi/8}_0 \frac{1}{4} \sin x <br />

    <br />
 [\frac{1}{4} \sin(4\theta)]^{\pi/8}_0<br />

    <br />
= \frac{1}{4} <br />
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Peleus View Post
    Yeah looking back it's quite silly of me. Brain snap.

    This could be just as silly, but I think it's better.

    Let 4\theta = x and  dx = 4d\theta

    <br />
\int^{\pi/8}_0 \cos x dx<br />

    <br />
\int^{\pi/8}_0 \frac{1}{4} \sin x <br />

    <br />
[\frac{1}{4} \sin(4\theta)]^{\pi/8}_0<br />

    <br />
= \frac{1}{4} <br />
    No need to go back to the original variable with definite integrals. Just substitute in the integral terminals as well. \theta = 0 \Rightarrow x = .... and \theta = \frac{\pi}{8} \Rightarrow x = .... and evaluate the new definite integral. Saves time and money.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integration by substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 3rd 2010, 11:12 PM
  2. Integration by Substitution Help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 1st 2009, 08:28 PM
  3. integration by substitution xe^x^2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 5th 2008, 10:48 PM
  4. integration by substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 6th 2008, 05:22 PM
  5. integration by substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 19th 2006, 03:46 PM

Search Tags


/mathhelpforum @mathhelpforum