Results 1 to 6 of 6

Math Help - integration of trigonometric functions

  1. #1
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161

    Thumbs up integration of trigonometric functions

    could someone check this for me?

    1.  \int (1+cos2x)sinx\ dx
    \int (1+cos^2x -sin^2x)sinx\ dx
     \int (2cos^2x)sinx\ dx
     \int 2cosx.sinx.cosx\ dx
     \int \frac{1}{2}sin2x.cosx\ dx
     u= \frac{1}{2}sin2x
     \frac{du}{dx} = cosx
     \int u\  dx
     =\frac{ u^2}{2} = \frac{(\frac{1}{2}sin2x)^2}{2} = \frac{\frac{1}{4}sin^22x}{2} = \frac{1}{8}sin^22x + c

    2.  \int sinx.cosx\ dx
     \int \frac{1}{2}sin2x\ dx
     = \frac{-1}{4}cos2x + c

    THANKS
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    Once you have int( 2cos^2(x)sin(x)dx)

    Let u = cos(x)

    then you have simply -2 int(u^2du)


    isn't correct since du/dx would be cos(2x)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,564
    Thanks
    1425
    Quote Originally Posted by differentiate View Post
    could someone check this for me?

    1.  \int (1+cos2x)sinx\ dx
    \int (1+cos^2x -sin^2x)sinx\ dx
     \int (2cos^2x)sinx\ dx Correct up to here.
     = -2\int{\cos^2{x}(-\sin{x})\,dx}

    Let u = \cos{x} so that \frac{du}{dx} = -\sin{x}.

    So the integral becomes

    -2\int{u^2\,\frac{du}{dx}\,dx}

     = -2\int{u^2\,du}

     = -\frac{2}{3}u^3 + C

     = -\frac{2}{3}\cos^3{x} + C.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,564
    Thanks
    1425
    Quote Originally Posted by differentiate View Post
    could someone check this for me?

    1.  \int (1+cos2x)sinx\ dx
    \int (1+cos^2x -sin^2x)sinx\ dx
     \int (2cos^2x)sinx\ dx
     \int 2cosx.sinx.cosx\ dx
     \int \frac{1}{2}sin2x.cosx\ dx
     u= \frac{1}{2}sin2x
     \frac{du}{dx} = cosx
     \int u\ dx
     =\frac{ u^2}{2} = \frac{(\frac{1}{2}sin2x)^2}{2} = \frac{\frac{1}{4}sin^22x}{2} = \frac{1}{8}sin^22x + c

    2.  \int sinx.cosx\ dx
     \int \frac{1}{2}sin2x\ dx
     = \frac{-1}{4}cos2x + c

    THANKS
    Q.2 is correct.

    Alternatively you can use a u substitution.

    \int{\sin{x}\cos{x}\,dx}.

    Let u = \sin{x} so that \frac{du}{dx} = \cos{x}.

    The integral becomes

    \int{u\,\frac{du}{dx}\,dx}

     = \int{u\,du}

     = \frac{1}{2}u^2 + C

     = \frac{1}{2}\sin^2{x} + C.


    It's not difficult to show that this is equivalent to what you posted.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161
    thank you. I get it now.

    so for this one:
    \int cosx.sin^4x\ dx
    u = sinx
     \frac{du}{dx} = cosx
    \int u^4\ du
     = \frac{u^5}{5} + c
     = \frac{sin^5x}{5} + c
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,564
    Thanks
    1425
    Quote Originally Posted by differentiate View Post
    thank you. I get it now.

    so for this one:
    \int cosx.sin^4x\ dx
    u = sinx
     \frac{du}{dx} = cosx
    \int u^4\ du
     = \frac{u^5}{5} + c
     = \frac{sin^5x}{5} + c
    Correct.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integration of trigonometric functions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 18th 2011, 09:17 AM
  2. Replies: 6
    Last Post: August 29th 2010, 05:23 PM
  3. Replies: 2
    Last Post: February 19th 2010, 10:37 AM
  4. Integration with Trigonometric Functions
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 15th 2010, 01:23 PM
  5. Integration of trigonometric functions
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: April 29th 2009, 09:21 AM

Search Tags


/mathhelpforum @mathhelpforum