Results 1 to 14 of 14

Math Help - Integration

  1. #1
    Member
    Joined
    Jan 2007
    Posts
    162

    Integration

    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by r_maths View Post
    you are correct up to the third to last line, you expanded incorrectly. the answer you should get is the one you said you should

    (1/2)(pi/4 - sin(2*(pi/4))/2) = pi/8 - (1/2)(1/2) = pi/8 - 1/4

    try the second question now
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2007
    Posts
    162
    Where did you get the bit in bold?
    (1/2)(pi/4 - sin(2*(pi/4))/2) = pi/8 - (1/2)(1/2) = pi/8 - 1/4


    From that, it gives 0. :S
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by r_maths View Post
    Where did you get the bit in bold?
    (1/2)(pi/4 - sin(2*(pi/4))/2) = pi/8 - (1/2)(1/2) = pi/8 - 1/4


    From that, it gives 0. :S
    i only did the first piece, the second function is zero so i forgot about it.

    sin[2*(pi/4)] = sin(pi/2) = 1

    so, (1/2)(pi/4 - sin(2*(pi/4))/2) = (1/2)(pi/4 - 1/2) = (1/2)(pi/4) - (1/2)(1/2)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2007
    Posts
    162
    O thanks alot. I can't believe I didn't see that.
    Im tired and I need sleep.
    Thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jan 2007
    Posts
    162
    Sorry, have to bump this thread again.
    Its question number 2.

    The full question:


    I've done part a.

    Answer


    Thanks
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by r_maths View Post
    Sorry, have to bump this thread again.
    Its question number 2.

    The full question:


    I've done part a.

    Answer


    Thanks
    ok, so the first question made no sense. it told you to find the answer by writing the answer...but anyway.

    for the second question, i would not use that. i would say that:

    cos^3x = cos^2x*cosx = (1 - sin^2x)cosx

    then use the substitution u = sinx, we'd end up integrating 1 - u^2 ...easy. but in the spirit of being as weird as this question is, let's do it like they said we should.

    we are told that

    cos(3x) = 4cos^3x - 3cosx
    => 4cos^3x = cos(3x) + 3cosx
    => cos^3x = (1/4)[cos(3x) + 3cosx]
    => int{cos^3x}dx = (1/4)*int{[cos(3x) + 3cosx]}dx
    you should be able to find the integral on the right without much problems...if you have problems, don't be afraid to say so
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,664
    Thanks
    298
    Awards
    1
    Quote Originally Posted by Jhevon View Post
    for the second question, i would not use that. i would say that:

    cos^3x = cos^2x*cosx = (1 - sin^2x)cosx
    The problem is dealing with cos(3x) not cos^3(x).

    For the record:
    cos(3x) = cos(2x + x) = cos(2x)cos(x) - sin(2x)sin(x)

    = (2cos^2(x) - 1)cos(x) - 2sin(x)cos(x)sin(x)

    = 2cos^3(x) - cos(x) - 2sin^2(x)cos(x)

    = 2cos^3(x) - cos(x) - 2[1 - cos^2(x)]cos(x)

    = 2cos^3(x) - cos(x) - 2cos(x) + 2cos^3(x)

    = 4cos^3(x) - 3cos(x)

    as advertised.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by topsquark View Post
    The problem is dealing with cos(3x) not cos^3(x).

    -Dan
    what do you mean? (b) says find int{cos^3x}dx
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,664
    Thanks
    298
    Awards
    1
    Quote Originally Posted by Jhevon View Post
    what do you mean? (b) says find int{cos^3x}dx
    Ah. I didn't look at that part.

    I think they want to use the identity as
    cos^3(x) = (1/4)cos(3x) + (3/4)cos(x)

    which is easier to integrate.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  11. #11
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by topsquark View Post
    Ah. I didn't look at that part.

    I think they want to use the identity as
    cos^3(x) = (1/4)cos(3x) + (3/4)cos(x)

    which is easier to integrate.

    -Dan
    which is what i said.

    and in regards to your second earlier post, this is what the question says.

    let the formula they have for cos(3x) be f(x)

    the question basically said.

    by writing cos(3x) as cos(3x) = f(x) show that cos(3x) = f(x)

    if that's not redundant i don't know what is
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Jan 2007
    Posts
    162
    Quote Originally Posted by Jhevon View Post
    ok, so the first question made no sense. it told you to find the answer by writing the answer...but anyway.
    It was one of the proof ones were you had to show/prove it just like what topsquark did...
    cos(3x) = cos(2x + x) = cos(2x)cos(x) - sin(2x)sin(x)

    = (2cos^2(x) - 1)cos(x) - 2sin(x)cos(x)sin(x)

    = 2cos^3(x) - cos(x) - 2sin^2(x)cos(x)

    = 2cos^3(x) - cos(x) - 2[1 - cos^2(x)]cos(x)

    = 2cos^3(x) - cos(x) - 2cos(x) + 2cos^3(x)

    = 4cos^3(x) - 3cos(x)
    Follow Math Help Forum on Facebook and Google+

  13. #13
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by r_maths View Post
    It was one of the proof ones were you had to show/prove it just like what topsquark did...
    yeah, i know what topsquark did was correct, i'm just saying that's not how the question was phrased. it should have said something like:

    "Prove that cos(3x) = 4cos^3x - 3cosx"

    and that's it.

    but since no one is seeing my point but me, let's just drop it. you have the correct answers
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,664
    Thanks
    298
    Awards
    1
    Quote Originally Posted by Jhevon View Post
    yeah, i know what topsquark did was correct, i'm just saying that's not how the question was phrased. it should have said something like:

    "Prove that cos(3x) = 4cos^3x - 3cosx"

    and that's it.

    but since no one is seeing my point but me, let's just drop it. you have the correct answers
    I agree: the first part of the question, at least, was badly worded.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 12:54 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 04:57 AM
  3. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  4. Replies: 2
    Last Post: February 19th 2010, 10:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum