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Math Help - 4 integration probs

  1. #1
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    4 integration probs

    If anybody can help me with at least one of these I would be grateful.

    1) the integral of (2e^x - 2e^-x)/[(e^x + e^-x)^2]

    2) the integral of (x^2)/(x^2 +4)

    3) The integral of (2x) divided by the square root of (4x-x^2)

    4) The integral from pi/4 to pi/2 of (cosx)/(1+(sinx)^2)
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by abcocoa View Post
    If anybody can help me with at least one of these I would be grateful.

    1) the integral of (2e^x - 2e^-x)/[(e^x + e^-x)^2]

    2) the integral of (x^2)/(x^2 +4)

    3) The integral of (2x) divided by the square root of (4x-x^2)

    4) The integral from pi/4 to pi/2 of (cosx)/(1+(sinx)^2)

    1) int{(2e^x - 2e^-x)/[(e^x + e^-x)^2]}dx

    We will do this by substitution.
    u = e^x + e^-x
    => du = e^x - e^-x dx

    so our integral becomes

    2*int{1/(u^2)}du
    = -2(u)^-1 + C
    = -2/(e^x + e^-x) + C

    2) int{(x^2)/(x^2 +4)}dx

    = int{(x^2 + 4 - 4)/(x^2 + 4)}dx
    = int{1 + 4/(x^2 + 4)}dx
    = int{1 + 4/[4((x/2)^2 + 1)]}dx
    = x - 2arctan(x/2) + C ............I skipped a step, to get here i used u = x/2 substitution.

    4) int{(cosx)/(1+(sinx)^2)}dx
    let u = sinx
    => du = cosx dx
    so our integral becomes
    int{1/(1 + u^2)}du
    = arctanu + C
    = arctan(sin(x)) + C
    evaluating between pi/4 and pi/2 we get

    arctan(sin(pi/2)) - arctan(sin(pi/4))
    = arctan1 - arctan(sqrt(2)/2)
    i don't rmember the exact value for these.


    3) Still working on this, I'm pretty sure we have to use trig substituion, which i'm a little rusty on, but there may be another way. I'm in a rush, so i'll type up to where i got and then leave the rest to you.

    int{2x/sqrt(4x - x^2)}dx
    = int{2x/sqrt[-(x^2 - 4x + (-2)^2 - (-2)^2)]}dx...............i'm completing the square here
    = int{2x/sqrt[4 - (x - 2)^2]}dx
    = int{2x/sqrt[2^2 - (x - 2)^2]}dx
    now we apply the trigonometric substitutions.
    we let x - 2 = 2sin(u)
    so dx = 2cos(u) du....and i'm late for class, catch you later
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  3. #3
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    Hello, abcocoa!

    Here's my version of #3 . . .



    . . . . . .2x dx
    3) .
    ----------
    . . . . √4x - x▓

    Complete the square and we have:

    . . . . . . . x dx
    . . 2
    --------------
    . . . . √4 - (x - 2)▓


    Let: .x - 2 .= .2Ěsinθ . . x .= .2Ěsinθ + 2 . . dx .= .2Ěcosθ dθ

    . . and the radical becomes: .2Ěcosθ


    . . . . . . . . . . . . (2Ěsinθ + 2)(2Ěcosθ dθ)
    Substitute: . 2
    ---------------------------- . = . 4 (sinθ + 1) dθ
    . . . . . . . . . . . . . . . . . 2Ěcosθ


    Can you finish it now?

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