Results 1 to 4 of 4

Math Help - Needs help on a tough question

  1. #1
    Junior Member
    Joined
    Jul 2006
    From
    Singapore
    Posts
    55

    Needs help on a tough question

    Question: Use the fact that 7 cos x - 4 sin x = (3/2)(cos x + sin x) + (11/2)(cos x - sin x) to find the exact value of (upper limit: 0.5?, lower limit: 0)?((7 cos x - 4 sin x)/(cos x + sin x)) dx.

    Answer: 3/4?

    I need help with the steps, thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello, margaritas!

    Edit: I got your correction . . . I'll edit my solution.


    Use the fact that: 7\cos x - 4\sin x \:= \:\frac{3}{2}(\cos x + \sin x) + \frac{11}{2}(\cos x - \sin x)

    to find the exact value of: . \int^{\frac{\pi}{2}}_0\frac{7\cos x - 4\sin x}{\cos x + \sin x}\,dx

    Answer: \frac{3}{4}\pi . . . Right!

    The function is: . \frac{7\cos x - 4\sin x}{\cos x + \sin x}\;= \;\frac{\frac{3}{2}(\cos x + \sin x) + \frac{11}{2}(\cos x - \sin x)}{\cos x + \sin x}

    . . = \;\frac{\frac{3}{2}(\cos x + \sin x)}{\cos x + \sin x} + \frac{\frac{11}{2}(\cos x - \sin x)}{\cos x + \sin x} =\;\frac{3}{2} + \frac{11}{2}\cdot\frac{\cos x - \sin x}{\cos x + \sin x}


    The integration is: . \frac{3}{2}\!\int\!\! dx + \frac{11}{2}\!\!\int\frac{\cos x - \sin x}{\cos x + \sin x}\,dx

    . . For the second integral, let  u = \cos x + \sin x\quad\Rightarrow\quad du = (\cos x - \sin x)\,dx

    . . then we have: . \int\frac{du}{u} \:=\:\ln|u| \:=\:\ln|\cos x + \sin x|

    Hence, we have: . \frac{3}{2}x + \frac{11}{2}\ln|\cos x + \sin x|\:\bigg]^{\frac{\pi}{2}}_0


    Evaluate: . \left[\frac{3}{2}\left(\frac{\pi}{2}\right) + \frac{11}{2}\ln\left|\cos \frac{\pi}{2} + \sin \frac{\pi}{2}\right|\right] -  \left[\frac{3}{2}\cdot0 + \frac{11}{2}\ln|\cos0 + \sin0|\right]

    . . = \;\left[\frac{3}{4}\pi + \frac{11}{2}\ln(0 + 1)\right] - \left[0 + \frac{11}{2}\ln(1 + 0)\right] \;= \;\boxed{\frac{3}{4}\pi}

    Last edited by Soroban; July 26th 2006 at 12:39 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    I recommend in this type of problem to use a Weierstrauss Substitution because it is a rational function of sine and cosine.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2006
    From
    Singapore
    Posts
    55
    Oops, all the '?'s should be pi's instead.

    But I think the solution by soroban will be helpful so thanks muchly!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tough vectors question...
    Posted in the Advanced Applied Math Forum
    Replies: 6
    Last Post: February 8th 2011, 10:30 AM
  2. Tough question
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: April 2nd 2010, 11:36 AM
  3. Tough (for me) arc length question
    Posted in the Calculus Forum
    Replies: 13
    Last Post: March 27th 2010, 04:05 PM
  4. tough question
    Posted in the Algebra Forum
    Replies: 15
    Last Post: September 4th 2009, 08:02 AM
  5. Very tough polynomial question.
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: December 1st 2008, 07:57 AM

Search Tags


/mathhelpforum @mathhelpforum