Page 2 of 2 FirstFirst 12
Results 16 to 19 of 19

Thread: U-Substitution Integrals

  1. #16
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by roshanhero View Post
    I am sorry but shouldnot there have been
    $\displaystyle

    \int{\frac{(1 + \tan{x})^3}{\cos^2{x}}\,dx} = \int{(1 + \tan{x})^3\sec^2{x}\,dx}
    $
    Yes.

    So the integral actually becomes

    $\displaystyle \int{u^3\,\frac{du}{dx}\,dx} = \int{u^3\,du}$

    $\displaystyle = \frac{1}{4}u^4 + C$

    $\displaystyle = \frac{1}{4}(1 + \tan{x})^4 + C$.
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Member
    Joined
    Aug 2008
    Posts
    249
    Quote Originally Posted by Prove It View Post
    The first two have been taken care of.

    for $\displaystyle \int{\frac{1}{1 + 4x^2}\,dx}$ you do not use a $\displaystyle u$ substitution. You use a trigonometric substitution.

    $\displaystyle \int{\frac{1}{1 + 4x^2}\,dx} = \int{\frac{1}{4\left(\frac{1}{4} + x^2\right)}\,dx}$

    $\displaystyle = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + x^2}\,dx}$.


    Now use the substitution $\displaystyle x = \frac{1}{2}\tan{\theta}$.

    Note that $\displaystyle \frac{dx}{d\theta} = \frac{1}{2}\sec^2{\theta}$.

    So $\displaystyle dx = \frac{1}{2}\sec^2{\theta}\,d\theta$.


    Therefore

    $\displaystyle = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + x^2}\,dx} = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + \left(\frac{1}{2}\tan{\theta}\right)^2}\,\frac{1}{ 2}\sec^2{\theta}\,d\theta}$

    $\displaystyle = \frac{1}{8}\int{\frac{1}{\frac{1}{4}(1 + \tan^2{\theta})}\,\sec^2{\theta}\,d\theta}$

    $\displaystyle = \frac{1}{2}\int{\frac{1}{1 + \tan^2{\theta}}\,\sec^2{\theta}\,d\theta}$

    $\displaystyle = \frac{1}{2}\int{\frac{1}{\sec^2{\theta}}\,\sec^2{\ theta}\,d\theta}$

    $\displaystyle = \frac{1}{2}\int{1\,d\theta}$

    $\displaystyle = \frac{1}{2}\theta + C$.


    Now remembering that $\displaystyle x = \frac{1}{2}\tan{\theta}$

    $\displaystyle 2x = \tan{\theta}$

    $\displaystyle \theta = \arctan{2x}$.


    So finally our answer is $\displaystyle \frac{1}{2}\arctan{2x} + C$.
    Actually I think doing a u-substitution for this problem would be much easier than doing a trig substitution.

    start by noticing that $\displaystyle \frac{1}{1 + 4x^2}$ = $\displaystyle \frac{1}{1+ (2x)^2}$

    let u = 2x and du = 2 dx. then you would get $\displaystyle \frac{1}{2}\$$\displaystyle \int{\frac{1}{1 + u^2}\,du}$

    then you would get the same answer $\displaystyle \frac{1}{2}\arctan{2x} + C$ with much less work.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by oblixps View Post
    Actually I think doing a u-substitution for this problem would be much easier than doing a trig substitution.

    start by noticing that $\displaystyle \frac{1}{1 + 4x^2}$ = $\displaystyle \frac{1}{1+ (2x)^2}$

    let u = 2x and du = 2 dx. then you would get $\displaystyle \frac{1}{2}\$$\displaystyle \int{\frac{1}{1 + u^2}\,du}$

    then you would get the same answer $\displaystyle \frac{1}{2}\arctan{2x} + C$ with much less work.
    That's only if you remember the rule for $\displaystyle \arctan$, which was found from trigonometric substitution...
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Member
    Joined
    Aug 2008
    Posts
    249
    Quote Originally Posted by Prove It View Post
    That's only if you remember the rule for $\displaystyle \arctan$, which was found from trigonometric substitution...
    I only used the definition of the derivative of arctan(x) along with a simple u-substitution to find this integral. no need for trig sub.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: Feb 16th 2011, 04:36 PM
  2. Integrals/Substitution help.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 19th 2010, 05:36 PM
  3. integrals by substitution
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Nov 11th 2008, 08:25 PM
  4. Using Substitution with Integrals
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Nov 11th 2008, 05:31 PM
  5. U - Substitution for integrals
    Posted in the Calculus Forum
    Replies: 13
    Last Post: Jun 13th 2008, 09:25 PM

Search Tags


/mathhelpforum @mathhelpforum