Results 1 to 6 of 6

Math Help - Integral....

  1. #1
    Senior Member polymerase's Avatar
    Joined
    May 2007
    From
    Sydney
    Posts
    267

    Integral....

    Find \int \frac{dx}{1+4sin^2x}

    thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    You can make varous subs to tackle this one.

    \int\frac{1}{1+4sin^{2}(x)}dx

    Let x=tan^{-1}(u), \;\ dx=\frac{1}{u^{2}+1}du

    After you make the subs and do the algebra, you should get:

    \int\frac{1}{5u^{2}+1}du

    Now, make the sub: u=\frac{tan(w)}{\sqrt{5}}, \;\ du=\frac{\sqrt{5}}{5cos^{2}(w)}dw

    Make the subs and it turns into a very friendly integral:

    \frac{1}{\sqrt{5}}\int{dw}

    Make the resubs and you're done.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member polymerase's Avatar
    Joined
    May 2007
    From
    Sydney
    Posts
    267
    Quote Originally Posted by galactus View Post
    You can make varous subs to tackle this one.

    \int\frac{1}{1+4sin^{2}(x)}dx

    Let x=tan^{-1}(u), \;\ dx=\frac{1}{u^{2}+1}du

    After you make the subs and do the algebra, you should get:

    \int\frac{1}{5u^{2}+1}du
    How did u simplify the expression after u made the substitution of x=tan^{-1}(u)? and how did you "see" this substitution?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    The simplification is just simple algebra. You shouldn't have any trouble since you're at this level of mathematics. As for 'seeing' the sub, it's a matter of doing them, I supoose.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Quote Originally Posted by polymerase View Post
    Find \int \frac{dx}{1+4sin^2x}
    \int {\frac{1}<br />
{{1 + 4\sin ^2 x}}\,dx} = \int {\frac{1}<br />
{{5\sin ^2 x + \cos ^2 x}}\,dx} = \int {\frac{{\sec ^2 x}}<br />
{{5\tan ^2 x + 1}}\,dx} .

    And this is an arctangent, so

    \int {\frac{1}<br />
{{1 + 4\sin ^2 x}}\,dx} = \frac{1}<br />
{{\sqrt 5 }}\arctan \left( {\sqrt 5 \tan x} \right) + k.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member polymerase's Avatar
    Joined
    May 2007
    From
    Sydney
    Posts
    267
    Quote Originally Posted by galactus View Post
    The simplification is just simple algebra. You shouldn't have any trouble since you're at this level of mathematics. As for 'seeing' the sub, it's a matter of doing them, I supoose.
    Thanks, i got it...i just fail to see at first that u=tan\:x! LOL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 08:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 03:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 02:52 PM
  4. Replies: 0
    Last Post: September 10th 2008, 08:53 PM
  5. Replies: 6
    Last Post: May 18th 2008, 07:37 AM

Search Tags


/mathhelpforum @mathhelpforum