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Math Help - Integral Problems; I'm stumped.

  1. #1
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    Integral Problems; I'm stumped.

    Integral (0, 2) of [(2)/(x+1) + (3)/(x-4)] dx





    Integral (0, pi) of cos^2(x) sin^2(x) dx


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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Kari View Post
    Integral (0, 2) of [(2)/(x+1) + (3)/(x-4)] dx





    Integral (0, pi) of cos^2(x) sin^2(x) dx

    For the first one try using this fact \int\frac{u'}{u}du=\ln|u|+C

    For the second one consider the following

    \begin{aligned}\cos^2(x)\sin^2(x)&=\left(\cos(x)\s  in(x)\right)^2\\<br />
&=\left(\frac{1}{2}\sin(2x)\right)^2\\<br />
&=\frac{\sin^2(2x)}{4}\\<br />
&=\frac{1-\cos(4x)}{8}\end{aligned}
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  3. #3
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    Thanks!
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  4. #4
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    Hello, I just finished the first problem I posted and was wondering if I could get it double checked? Thanks in advance!

    Integral (0,2) of [(2/(x+1) + (3/(x-4)]dx

    Integral (0,2) 2/(x+1) dx + Integral (0,2) 3/(x-4) dx

    Let u = x+1; Let u = x-4;
    du/dx = 1 du/dx = 1
    du=dx du=dx

    Integral (0,2) 2/u du + Integral (0,2) 3/u du

    2*Integral 1/u du + 3*Integral 1/u du
    2lnu + 3lnu
    2ln(x+1) + 3ln(x-4)

    Plugging in the intervals:


    [2ln(2+1) + 3ln(2-4)] - [2ln(1) + 3ln(-4)]

    [ln(3)^2 + ln(-2)^3] - [ln(1)^2 + ln(-4)^3]

    [ln(9) + ln(-8)] - [ln(1) + ln(-64)]

    ln(-72) - ln(-64)

    ln(-72/-64)

    ln (9/8) = 0.0511525224
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  5. #5
    Math Engineering Student
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    Quote Originally Posted by Kari View Post

    [ln(9) + ln(-8)] - [ln(1) + ln(-64)]

    ln(-72) - ln(-64)

    ln(-72/-64)

    ln (9/8) = 0.0511525224
    May be you get the correct answer but your algebra is not correct, since, logarithms are defined for positive values, not for negative ones, hence, you should include the absolute value in each one of them.
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