Results 1 to 3 of 3

Math Help - Integrals!

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

    Integrals!

    \int \frac{sin\;3x}{1+cos^2\;x}\;dx

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Quote Originally Posted by polymerase View Post
    \int \frac{sin\;3x}{1+cos^2\;x}\;dx
    You can easily show that \sin 3x = 3\sin x - 4\sin ^3 x.

    Now try to find an arctangent:

    \int {\frac{{\sin 3x}}<br />
{{1 + \cos ^2 x}}\,dx}  = \int {\frac{{3\sin x - 4\sin ^3 x}}<br />
{{1 + \cos ^2 x}}\,dx}  = \int {\frac{{\left( {8\sin x - 4\sin ^3 x} \right) - 5\sin x}}<br />
{{1 + \cos ^2 x}}\,dx} .

    Note that 8\sin x - 4\sin ^3 x = 4\sin x\left( {2 - \sin ^2 x} \right) = 4\sin x\left( {1 + \cos ^2 x} \right).

    The rest follows.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    779
    Hello, polymerase!

    I found an approach . . . but there must be a better way!


    \int \frac{\sin3x}{1+\cos^2\!x}\,dx
    Multiple-angle identity: . \sin3\theta \:=\:3\sin\theta - 4\sin^3\!\theta

    We have: . \int\frac{3\sin x-4\sin^3\!x}{1 + (1-\sin^2\!x)}\,dx \;=\;\int\frac{3\sin x-4\sin^3\!x}{2-\sin^2\!x}\,dx \;=\;\int\frac{4\sin^3\!x-3\sin x}{\sin^2\!x-2}\,dx

    Long division: . \int\left(4\sin x + \frac{5\sin x}{\sin^2\!x-2}\right)dx \;=\;\int\left(4\sin x + \frac{5\sin x}{(1-\cos^2\!x) - 2}\right)dx

    . . and we have: . 4\int\sin x\,dx - 5\int\frac{\sin x}{1 + \cos^2\!x}\,dx \;=\;-4\cos x - 5\int\frac{\sin x}{1+\cos^2\!x}\,dx


    Let: . u \:= \:\cos x\quad\Rightarrow\quad du \:=\: -\sin x\,dx

    Substitute: . -4\cos x + 5\int\frac{du}{1+u^2} \;=\;-4\cos x + 5\arctan u + C


    Back-substitute: . \boxed{-4\cos x + 5\arctan(\cos x) + C}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Contour Integrals (to Evaluate Real Integrals)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: January 17th 2011, 10:23 PM
  2. Replies: 1
    Last Post: December 6th 2009, 08:43 PM
  3. Integrals : 2
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 24th 2009, 08:40 AM
  4. Integrals and Indefinite Integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 9th 2009, 05:52 PM
  5. integrals Help please
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 8th 2008, 07:16 PM

Search Tags


/mathhelpforum @mathhelpforum