Results 1 to 3 of 3

Math Help - Help needed please...

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    4

    Help needed please...

    Have been given this question and am strugling abit.
    Any help appreciated:

    Find the work done (W) by a mechanism between the angular displacement (x) of 0 and pie/2, by intergrating the following equation:

    2cos(2x) + 3sin(3x)

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hello
    Quote Originally Posted by oldskool_89 View Post
    Find the work done (W) by a mechanism between the angular displacement (x) of 0 and pie/2, by intergrating the following equation:

    2cos(2x) + 3sin(3x)
    W=\int_{0}^{\frac{\pi}{2}}2\cos(2x)+3\sin(3x)\,\ma  thrm{d}x=\int_{0}^{\frac{\pi}{2}}2\cos(2x)\,\mathr  m{d}x+\int_{0}^{\frac{\pi}{2}}3\sin(3x)\,\mathrm{d  }x

    As \frac{\mathrm{d}}{\mathrm{d}x}(2x)=2, one can writes 2\cos(2x)=u'(x)\cdot \cos(u(x)) for u(x)=2x. Can you find an anti-derivative of this function ? (think about the chain rule : (v\circ u)'=u'\cdot v'\circ u. You know what is u and just need to find v)

    The same idea applies for the other integral.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2007
    Posts
    162
    2\cos (2x)+3\sin (3x)

    Integration, cos goes to sin and sin goes to -cos.

    Also, remember the chain rule for inside the brackets, differentiate the bracket and divide by what you get.

    \int{2\cos (2x)}=\frac{2\sin (2x)}{2}=\sin (2x) - (cancel the 2s)

    \int{3\sin (3x)=-\frac{3\cos (3x)}{3}}=-\cos (3x) - (cancel the 3s)

    Answer: \int_{0}^{\frac{\pi }{2}}{2\cos (2x)+3\sin (3x)=\left[ \sin (2x)-\cos (3x) \right]_{0}^{\frac{\pi }{2}}}
    Last edited by r_maths; May 21st 2008 at 04:17 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A little bit of help needed please!
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 4th 2009, 10:19 AM
  2. Help needed...thanks
    Posted in the Algebra Forum
    Replies: 6
    Last Post: April 15th 2009, 05:23 AM
  3. Help needed please
    Posted in the Algebra Forum
    Replies: 10
    Last Post: January 20th 2009, 07:40 PM
  4. more help needed
    Posted in the Algebra Forum
    Replies: 1
    Last Post: July 21st 2008, 09:04 AM
  5. Help needed please!
    Posted in the Calculus Forum
    Replies: 10
    Last Post: April 27th 2008, 02:16 PM

Search Tags


/mathhelpforum @mathhelpforum