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Math Help - Problem with integration

  1. #1
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    Problem with integration

    Hello, I am currently working on a problem that requires the integration of an expression that I can't seem to figure out. The problem goes:

    Integral from 0 to 2 of: (((t^2)+2t+2)/((1+t)^4))^1/2

    So far, I know that I have to use u-substitution and I set u = 1+t, so du = dt. Now, i have

    Integral of: ((u^2 + 1)^1/2)/(u^2)

    but after this, I don't know how to continue integrating. Could anyone please help me?????
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  2. #2
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    Quote Originally Posted by collegestudent321 View Post
    Hello, I am currently working on a problem that requires the integration of an expression that I can't seem to figure out. The problem goes:

    Integral from 0 to 2 of: (((t^2)+2t+2)/((1+t)^4))^1/2

    So far, I know that I have to use u-substitution and I set u = 1+t, so du = dt. Now, i have

    Integral of: ((u^2 + 1)^1/2)/(u^2)

    but after this, I don't know how to continue integrating. Could anyone please help me?????
    Try u = \tan \theta.
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    so use u = tan (t) instead of (1+t)??
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    Quote Originally Posted by collegestudent321 View Post
    so use u = tan (t) instead of (1+t)??
    No No. Your first substitution is

    t + 1 = u, your second is u = \tan \theta.
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    ok, so now i have that u = tan(theta)
    and now it is:
    integral of : ((tan(theta)^2)+1)^1/2) / (tan(theta)^2) which equals:

    integral of: sec(theta) / ((sec(theta)^2) - 1)

    is that right so far? and if it is, where do i go now?
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  6. #6
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    Quote Originally Posted by collegestudent321 View Post
    ok, so now i have that u = tan(theta)
    and now it is:
    integral of : ((tan(theta)^2)+1)^1/2) / (tan(theta)^2) which equals:

    integral of: sec(theta) / ((sec(theta)^2) - 1)

    is that right so far? and if it is, where do i go now?
    I think you're missing a du however with the tan subs., it's going to be a long process. Let's try another substitution.

    u = \sinh \theta
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    ok, now i'm completely lost. I tried the same process as before and now i have:

    integral of: cosh(t) / sinh(t)^2 dt

    where do i go from here?
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  8. #8
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    wait, can this be equal to :

    integral of : coth(t)*csch(t) ???

    and then use integration by parts?
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  9. #9
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    OK. Let me put in some steps to help clear this up.If \int \frac{\sqrt{u^2+1}}{u^2}\, du if u = \sinh \theta then \int \frac{\sqrt{\sinh^2 \theta+1}}{\sinh^2 \theta}\, \cosh \theta d \theta = \int \frac{\cosh^2 \theta}{\sinh^2 \theta} d \theta

    so
    \int \frac{\sinh^2 \theta +1}{\sinh^2 \theta} d \theta = \int ( 1 + \text{csch}^2 \theta ) d \theta = \theta - \coth \theta + c. Now backsubstitute.
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  10. #10
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    OHHHHHHH. wow i got it. the answer is:

    ln (u + ((u^2)+1)) - ((1-(u^2))^1/2)/(u)

    (and then substitute u with (t+1)

    Thank you for your help. I REALLY appreciate it.
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  11. #11
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    calculus with parametric curves

    actually, if you don't mind, there is one other problem that i've been having. I'm sorry to bug you so much.

    I have to prove that if x = asin(theta) and y = bcos(theta), show that the total length of the ellipse is:

    L = 4a * integral (from 0 to pi/2) of: 1 - ((e^2)sin(theta)^2))^1/2 (d(theta))


    I am almost there, but i came across this part:

    L = integral of: (a^2) - (c^2)(sin(theta)^2) and i'm not sure how to make the connection to find e^2. Any ideas????
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  12. #12
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    Quote Originally Posted by collegestudent321 View Post
    actually, if you don't mind, there is one other problem that i've been having. I'm sorry to bug you so much.

    I have to prove that if x = asin(theta) and y = bcos(theta), show that the total length of the ellipse is:

    L = 4a * integral (from 0 to pi/2) of: 1 - ((e^2)sin(theta)^2))^1/2 (d(theta))


    I am almost there, but i came across this part:

    L = integral of: (a^2) - (c^2)(sin(theta)^2) and i'm not sure how to make the connection to find e^2. Any ideas????
    L = 4 \int_0^{\pi/2} \sqrt{x'^2+y'^2} d \theta = 4\int_0^{\pi/2} \sqrt{ a^2 \cos^2 \theta + b^2 \sin^2 \theta} d \theta =

     4a\int_0^{\pi/2} \sqrt{\cos^2 \theta + \frac{b^2}{a^2} \sin^2 \theta} d \theta = 4a\int_0^{\pi/2} \sqrt{1 - \left(1 - \frac{b^2}{a^2}\right) \sin^2 \theta} d \theta

    Now b^2 = a^2 - c^2 and e = \frac{c}{a} so using these we get

     4a\int_0^{\pi/2} \sqrt{1 - \left(1 - \frac{b^2}{a^2}\right) \sin^2 \theta} d \theta = 4a\int_0^{\pi/2} \sqrt{1 - e^2\sin^2 \theta} d \theta
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  13. #13
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    oh, so was i supposed to know that e = c/a or is that derived from the problem???
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  14. #14
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    Quote Originally Posted by collegestudent321 View Post
    oh, so was i supposed to know that e = c/a or is that derived from the problem???
    The eccentricity e is usually defined that way - have you not seen it?
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  15. #15
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    ohhh, ok. haha wow, i was thinking the other e. ok, that makes sense, yea i have seen that before, i just hadn't applied it. thank you!!!
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