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Math Help - Help with these integrals:

  1. #1
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    Help with these integrals:

    1.)
    Answer Key:

    Is it necessary to use integration by parts here? cause it wasnt discussed in this lesson yet

    2.)
    Answer key:


    please take your time..
    and thanks a kazillions
    Last edited by ^_^Engineer_Adam^_^; March 23rd 2007 at 07:48 PM. Reason: Just adding
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    1.)
    Answer Key:

    Is it necessary to use integration by parts here? cause it wasnt discussed in this lesson yet

    2.)
    Answer key:


    please take your time..
    and thanks a kazillions
    The second integratoin has me stumped right now. I'll try coming back to it later, but for now, I can help you with the first integration:

    1. INT(Sqrt(3 - 2x)x^2) dx
    This can be solved without integration by parts. You might find it hard to believe, but we can actually solve this by u substitution:

    Let u = 3 - 2x ... Why would I do that? You'll see .
    du = -2dx --> dx = -du/2
    Before I go on, let's solve the u = 3 - 2x for x (we'll need this in a second): x = 1/2(3 - u)

    Ok, now watch my slight of hand as I do some math-magic on this integration...

    INT(Sqrt(u)x^2) -du/2 ... Now you see an x^2,
    -1/2 INT(Sqrt(u)[1/2(3 - u)]^2) du ... Now you don't. ... Guess where it went. (Hint: x = 1/2(3 - u))

    From here, I think you can solve it the rest of the way, just multiply out the [1/2(3 - u)]^2 then multiply this to the Sqrt(u).
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    1.)
    Answer Key:

    Is it necessary to use integration by parts here? cause it wasnt discussed in this lesson yet

    2.)
    Answer key:


    please take your time..
    and thanks a kazillions
    This second integration throws a giant curve ball at you. I've figured out how to do it, it's just that the way to do it is a little tricky.

    INT (tan(2x) + cot(2x))^2 dx ... let's turn these into sines and cosines
    INT (sin(2x)/cos(2x) + cos(2x)/sin(2x))^2 ... now, get common denominators to add the fractions
    INT [((sin(2x))^2 + (cos(2x))^2)/(sin(2x)cos(2x)]^2 dx
    INT 1/((sin2x)^2(cos2x)^2) dx

    Now here's the catch, we can use (sin2x)^2 + (cos2x)^2 to replace the 1 in the numerator.

    You might be thinking, why use that identity in reverse when we already used it to get rid of (sin2x)^2 + (cos2x)^2?

    The reason we had to get rid of the (sin2x)^2 + (cos2x)^2 a few steps before is because that whole term was being raised to the second power. If we had multiplied all that out, the problem would get really messy, but there's nothing wrong with replacing 1 with (sin2x)^2 + (cos2x)^2, so that's what we'll do.

    INT ((sin2x)^2 + (cos2x)^2)/((sin2x)^2(cos2x)^2) dx

    Now let's split up this fraction again. We should get:

    INT 1/(cos2x)^2 + 1/(sin2x)^2 dx
    INT (sec2x)^2 + (csc2x)^2 ... This we can integrate.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    This second integration throws a giant curve ball at you. I've figured out how to do it, it's just that the way to do it is a little tricky.

    INT (tan(2x) + cot(2x))^2 dx ... let's turn these into sines and cosines
    INT (sin(2x)/cos(2x) + cos(2x)/sin(2x))^2 ... now, get common denominators to add the fractions
    INT [((sin(2x))^2 + (cos(2x))^2)/(sin(2x)cos(2x)]^2 dx
    INT 1/((sin2x)^2(cos2x)^2) dx

    Now here's the catch, we can use (sin2x)^2 + (cos2x)^2 to replace the 1 in the numerator.

    You might be thinking, why use that identity in reverse when we already used it to get rid of (sin2x)^2 + (cos2x)^2?

    The reason we had to get rid of the (sin2x)^2 + (cos2x)^2 a few steps before is because that whole term was being raised to the second power. If we had multiplied all that out, the problem would get really messy, but there's nothing wrong with replacing 1 with (sin2x)^2 + (cos2x)^2, so that's what we'll do.

    INT ((sin2x)^2 + (cos2x)^2)/((sin2x)^2(cos2x)^2) dx

    Now let's split up this fraction again. We should get:

    INT 1/(cos2x)^2 + 1/(sin2x)^2 dx
    INT (sec2x)^2 + (csc2x)^2 ... This we can integrate.
    YOU ARE THE MAN ECMATHGEEK!!!!! by the way, what does the ec stand for?
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  5. #5
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Jhevon View Post
    YOU ARE THE MAN ECMATHGEEK!!!!! by the way, what does the ec stand for?
    Thank you.

    The real reason I have the letters e and c in my handle is kind of lame, so let's just say I really like the numbers e and c.

    I use the "ec" with "MathGeek" because "MathGeek" alone is usually taken in most things I try to sign up for, whereas "ecMathGeek" never is.
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  6. #6
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    1.)
    Answer Key:

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  7. #7
    Math Engineering Student
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    For the second one just expand it and then it's done...
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  8. #8
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Krizalid View Post
    For the second one just expand it and then it's done...
    ... I tried expanding it, it became a giant mess.

    Ok, maybe not a giant mess, but it wasn't easy to integrate without some special trig-identity substitutions.
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  9. #9
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    Second one:

    int[(tan(2x)+cot(2x))^2]dx =

    int[ (sin(2x)/cos(2x) + cos(2x)/sin(2x))^2]dx =

    int[ ((sin(2x))^2 + (cos(2x))^2))/ cos(2x)sin(2x))^2 ] =

    int[(1/sin(2x)cos(2x))^2]dx (i used sin(2x)^2 + cos(2x)^2 = 1 identity)

    = subtitue 2x=t ==> dx = dt/2

    = 1/2int[ (1/costsint)^2] dt (2sinxcosx = sin2x ==> sinxcosx = 1/2sin2x)

    = 1/2int[ (2/sin2t)^2 ]dt = 2int[1/sin(2t)^2]dt = -2cot(2t)/2 = -cot4x

    i guess you can open -cot4x for the answer they request using identities
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  10. #10
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    Thanks a lot rookies
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  11. #11
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    Quote Originally Posted by ecMathGeek View Post
    ... I tried expanding it, it became a giant mess.
    I don't think so, look



    As desired...
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  12. #12
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Krizalid View Post
    I don't think so, look



    As desired...
    lol. Nicely done. I was intent on changing them to sines and cosines and doing so made them a giant mess. I did demonstrate that it could be done with sines and cosines but thank you for clarifying that it wasn't needed. I didn't even notice that that could be done.
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  13. #13
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    Hello, ^_^Engineer_Adam^_^!

    The second one can be done without switching functions . . .


    [tan(2x) + cot(2x)]˛ dx

    [tan(2x) + cot(2x)]˛ .= .tan˛(2x) + 2ˇtan(2x)ˇcot(2x) + cot˛(2x)

    . . . . . . . . . . . . . . .= .tan˛(2x) + 2 + cot˛(2x)

    . . . . . . . . . . . . . . .= .[sec˛(2x) - 1] + 2 + [csc˛(2x) - 1]

    . . . . . . . . . . . . . . .= .sec˛(2x) + csc˛(2x)


    The integral becomes: .
    [sec˛(2x) + csc˛(2x)] dx . = . ˝ˇtan(2x) - ˝ˇcot(2x) + C

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