Results 1 to 8 of 8

Math Help - integrating sinx(1+cosx)

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    12

    Smile integrating sinx(1+cosx)

    hey .
    I need some help in integrating sinx(1+cosx) I've tried doing integration by parts but got in a big mess.
    Help would be greatly appreciated

    thankyou
    ellie
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by elliegurl View Post
    hey .
    I need some help in integrating sinx(1+cosx) I've tried doing integration by parts but got in a big mess.
    Help would be greatly appreciated

    thankyou
    ellie
    Have you got some working to show?

    I would expand to get  \int \sin x - \cos x \sin x ~dx = \int \sin x ~dx-\int \cos x \sin x ~dx

    u-sub for 2nd part.
    Last edited by pickslides; May 2nd 2010 at 03:21 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Quote Originally Posted by elliegurl View Post
    \int{\sin{x}(1+\cos{x})}\;{dx}
    By substitution:
    Spoiler:
    Let u = \cos{x}, then \dfrac{du}{dx} = -\sin{x} \Rightarrow {dx} = -\dfrac{du}{\sin{x}}.
    So \int{\sin{x}(1+\cos{x})}\;{dx} = -\int\dfrac{\sin{x}(1+u)}{\sin{x}}\;{du} = -\int{\left(1+u\right)}\;{du} = -\left(u+\dfrac{u^2}{2}\right)+C

    Therefore \int{\sin{x}(1+\cos{x})}\;{dx} = \boxed{-\dfrac{1}{2}\cos^2{x}-\cos{x}+C}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by elliegurl View Post
    hey .
    I need some help in integrating sinx(1+cosx) I've tried doing integration by parts but got in a big mess.
    Help would be greatly appreciated

    thankyou
    ellie
    Another way:

    \int \sin(x) + \int \sin(x) \cos(x)

    \sin(x) \cos(x) = \frac{1}{2} \sin(2x)


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



    Spoiler:
    -\cos(x) - \frac{1}{4} \cos(2x) + k

    which is equivalent to TheCoffeeMachine's answer. I have used k as the constant of integration because I have shown that the two answers are equivalent below

     = -\cos(x) - \frac{1}{4} (2\cos^2(x) -1) + k

    = -\frac{1}{2}\cos^2(x) - \cos(x) + \frac{1}{4}+k

    = -\frac{1}{2}\cos^2(x) - \cos(x) + C where C=\frac{1}{4}+k
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2008
    Posts
    12

    thankyou

    Hey thanks for all your help. the limits were pi and 0 and when i sub them in to both of your answers I get 2 which is right but in the markscheme the integral they have is
    -0.5(1+cosx)^2 . I cant see how they have got that. Im a bit confused
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    Quote Originally Posted by elliegurl View Post
    Hey thanks for all your help. the limits were pi and 0 and when i sub them in to both of your answers I get 2 which is right but in the markscheme the integral they have is
    -0.5(1+cosx)^2 . I cant see how they have got that. Im a bit confused
    Put (1+cosx) = t
    then -sinx*dx = dt
    So the integration becomes
    -intg(t)dt. That is equal to
    -0.5t^2
    = -0.5(1+cosx)^2
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2008
    Posts
    12
    Quote Originally Posted by sa-ri-ga-ma View Post
    Put (1+cosx) = t
    then -sinx*dx = dt
    So the integration becomes
    -intg(t)dt. That is equal to
    -0.5t^2
    = -0.5(1+cosx)^2


    thankyou so much
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member slovakiamaths's Avatar
    Joined
    Apr 2010
    From
    India
    Posts
    41
    For Solution see the attachment
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. sinx+cosx=0
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: July 24th 2011, 08:21 AM
  2. sin a sinx+ b cosx
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 23rd 2009, 08:48 AM
  3. Verify that √((1-cosx)/(1+cosx)) = (1-cosx)/|sinx|
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: December 10th 2008, 09:39 PM
  4. cosx and sinx when x= 0
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 1st 2008, 06:26 PM
  5. cosx-sinx
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: July 29th 2008, 02:26 PM

Search Tags


/mathhelpforum @mathhelpforum