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Math Help - Integration by Parts and Arclength

  1. #1
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    Integration by Parts and Arclength

    Back again! This time I have five problems that I have completed that, if anyone would be so kind, I would like to have looked over and corrected if they are wrong. Thank you very very much in advance!

    Problem 1. \int^{e^4}_e\frac{1}{xln(x)} dx

    My Solution

    u = ln(x), du = \frac{1}{x} dx, dx = x du

    \int\frac{1}{xu}x du = \int\frac{1}{u} du = ln(u) + C = [ln(lnx)]^{e_4}_e = ln[ln( e^4)] - ln[ln(e)] = ln[ln( e^4)] - 0 = ln[ln( e^4)]

    Problem 2. \intxsin4x dx

    My Solution


    u = x, du = dx, dv = sin4x, v = \frac{-1}{4}cos4x

    x(\frac{-1}{4}cos4x) - \int\frac{-1}{4}cos4x dx = \frac{-x}{4}cos4x + \frac{1}{4}\int cos4x dx = \frac{-x}{4}cos4x + \frac{1}{4}(\frac{1}{4}sin4x) + C = \frac{-x}{4} + \frac{1}{16}sin4x + C = \frac{-4x + sin4x}{16} + C

    Problem 3. \int\frac{7x}{4 + 9x^2} dx

    My Solution

    u = 4 + 9 x^2, du = 18x dx, dx = \frac{1}{18}x du

    \int\frac{7x}{u}\frac{1}{18x} du = \int\frac{7}{18u} du = \frac{7}{18}\int\frac{1}{u} du = \frac{7}{18}ln|u| + C = \frac{7}{18}ln(4 + 9x^2) + C

    Problem 4. \int\frac{5}{3x^2 - 12x + 24} dx

    My Solution

    \frac{1}{3}\int\frac{5}{x^2 - 4x + 8} dx = \frac{5}{3}\int\frac{1}{x^2 - 4x + 8} dx = \frac{5}{3}\int\frac{1}{(x + 4)^2 + 4} dx = \frac{5}{3}[\frac{1}{2}arctan(\frac{x + 4}{2})] + C = \frac{5}{6}arctan(\frac{x + 4}{2}) + C

    Problem 5. Calculate the length along the curve x = cos^3(t), y = sin^3(t), 0 \leq t \leq \frac{\pi}{2}

    My Solution (Not so sure about this one.)

    \frac{dx}{dt} = -cos^2(t)sin(t), \frac{dy}{dt} = sin^2(t)cos(t)

    L = \int^{\frac{\pi}{2}}_0\sqrt{(-cos^2(t)sin(t))^2 + (sin^2(t)cos(t))^2} dt = \int^{\frac{\pi}{2}}_0\sqrt{cos^4(t)sin^2(t) + sin^4(t)cos^2(t)} dt = \int^{\frac{\pi}{2}}_0\sqrt{cos^2(t)sin^2(t)(cos^2  (t) + sin^2(t)} dt = \int^{\frac{\pi}{2}}_0\sqrt{cos^2(t)sin^2(t)} dt = \int^{\frac{\pi}{2}}_0cos(t)sin(t) dt

    u = cos(t), du = -sin(t) dt, dt = \frac{-1}{sin(t)} du

    -\int usin(t)\frac{1}{sin(t)} du = -\intu du = \frac{-u^2}{2} + C = -[\frac{cos^2(t)}{2}]^{\frac{\pi}{2}}_0 = -[0 - \frac{1}{2}] = \frac{1}{2}

    Once again, thank you very much in advance.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Good Work

    only thing:

    1) ln(ln(e^4)) = ln(4)
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  3. #3
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    Quote Originally Posted by mturner07 View Post
    Problem 4. \int\frac{5}{3x^2 - 12x + 24} dx

    My Solution

    \frac{1}{3}\int\frac{5}{x^2 - 4x + 8} dx = \frac{5}{3}\int\frac{1}{x^2 - 4x + 8} dx = \frac{5}{3}\int\frac{1}{(x + 4)^2 + 4} dx = \frac{5}{3}[\frac{1}{2}arctan(\frac{x + 4}{2})] + C = \frac{5}{6}arctan(\frac{x + 4}{2}) + C

    \textcolor{red}{\frac{1}{x^2-4x+8} = \frac{1}{(x-2)^2 + 4}}<br />
    ...
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  4. #4
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    Thank you. My new answer is \frac{5}{6}arctan(\frac{x-2}{2}) + C Is this correct?
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  5. #5
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    In the first problem notice that the limits of integration can also go through the u-substitution process, and you don't need to return to terms of x. Since e and e^4 are in terms of x, and u = ln(x), then:

    \int_e^{e^4}\frac{1}{xln(x)} dx = \int_{ln(e)}^{ln(e^4)}\frac{1}{u} du = \int_{1}^{4}\frac{1}{u} du = ln(u)_{1}^{4} = ln(4) - ln(1) = ln(4)

    It isn't always easy or better to convert your limits of integration to the terms of your substitution, but sometimes it does simplify things.


    Brian
    (Who is only about a semester past this point, so I hope what I've said is all correct!)
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  6. #6
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    Quote Originally Posted by mturner07 View Post
    Back again! This time I have five problems that I have completed that, if anyone would be so kind, I would like to have looked over and corrected if they are wrong. Thank you very very much in advance!

    Problem 1. \int^{e^4}_e\frac{1}{xln(x)} dx

    My Solution

    u = ln(x), du = \frac{1}{x} dx, dx = x du

    \int\frac{1}{xu}x du = \int\frac{1}{u} du = ln(u) + C = [ln(lnx)]^{e_4}_e = ln[ln( e^4)] - ln[ln(e)] = ln[ln( e^4)] - 0 = ln[ln( e^4)]
    It is far simpler just to use substitution. Let u= ln(x) so that du= dx/x. When x= e, u= ln(e)= 1. When x= e^4, [tex]u= ln(e^4)= 4. The integral becomes \int_1^4 \frac{1}{u}du

    Problem 2. \intxsin4x dx

    My Solution


    u = x, du = dx, dv = sin4x, v = \frac{-1}{4}cos4x

    x(\frac{-1}{4}cos4x) - \int\frac{-1}{4}cos4x dx = \frac{-x}{4}cos4x + \frac{1}{4}\int cos4x dx = \frac{-x}{4}cos4x + \frac{1}{4}(\frac{1}{4}sin4x) + C = \frac{-x}{4} + \frac{1}{16}sin4x + C = \frac{-4x + sin4x}{16} + C

    Problem 3. \int\frac{7x}{4 + 9x^2} dx

    My Solution

    u = 4 + 9 x^2, du = 18x dx, dx = \frac{1}{18}x du

    \int\frac{7x}{u}\frac{1}{18x} du = \int\frac{7}{18u} du = \frac{7}{18}\int\frac{1}{u} du = \frac{7}{18}ln|u| + C = \frac{7}{18}ln(4 + 9x^2) + C

    Problem 4. \int\frac{5}{3x^2 - 12x + 24} dx

    My Solution

    \frac{1}{3}\int\frac{5}{x^2 - 4x + 8} dx = \frac{5}{3}\int\frac{1}{x^2 - 4x + 8} dx = \frac{5}{3}\int\frac{1}{(x + 4)^2 + 4} dx = \frac{5}{3}[\frac{1}{2}arctan(\frac{x + 4}{2})] + C = \frac{5}{6}arctan(\frac{x + 4}{2}) + C

    Problem 5. Calculate the length along the curve x = cos^3(t), y = sin^3(t), 0 \leq t \leq \frac{\pi}{2}

    My Solution (Not so sure about this one.)

    \frac{dx}{dt} = -cos^2(t)sin(t), \frac{dy}{dt} = sin^2(t)cos(t)

    L = \int^{\frac{\pi}{2}}_0\sqrt{(-cos^2(t)sin(t))^2 + (sin^2(t)cos(t))^2} dt = \int^{\frac{\pi}{2}}_0\sqrt{cos^4(t)sin^2(t) + sin^4(t)cos^2(t)} dt = \int^{\frac{\pi}{2}}_0\sqrt{cos^2(t)sin^2(t)(cos^2  (t) + sin^2(t)} dt = \int^{\frac{\pi}{2}}_0\sqrt{cos^2(t)sin^2(t)} dt = \int^{\frac{\pi}{2}}_0cos(t)sin(t) dt

    u = cos(t), du = -sin(t) dt, dt = \frac{-1}{sin(t)} du

    -\int usin(t)\frac{1}{sin(t)} du = -\intu du = \frac{-u^2}{2} + C = -[\frac{cos^2(t)}{2}]^{\frac{\pi}{2}}_0 = -[0 - \frac{1}{2}] = \frac{1}{2}

    Once again, thank you very much in advance.
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