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Help with Trigonometric Integrals

  1. Mar 28, 2014 #1
    Could someone help me with these two problems? I've been at them for an hour, but have very little clue how to go about solving either of them.

    1. The problem statement, all variables and given/known data

    1)∫ 6 csc^3 (x) cot x dx

    2. Relevant equations

    3. The attempt at a solution

    6 ∫ csc^3 (x) dx) / tan x

    csc^3 / tan x = csc^3 cot x

    cot^2 x = csc^2 x - 1
    csc^2 x = cot^2 x + 1

    csc x cot x (cot^2 x + 1)

    u = csc x
    du = - csc x cot x dx

    1. The problem statement, all variables and given/known data

    2)Find the length of the curve: y = ln(csc x), π/4 <= x <= π/2

    2. Relevant equations


    3. The attempt at a solution

    L = ∫sqrt(1 + (d(ln(csc x))^2) dx from x = π/4 to π/2

    d(ln(csc x)/dx = -cot x
    (1 + cot x^2) = csc^2

    L = ∫ csc x dx from x = π/4 to π/2
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 28, 2014 #2

    SammyS

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    The two problems are not all that closely related.

    It's generally better to post these in separate threads.


    For the first problem.

    The substitution u = csc(x) is a good choice.

    As you noted, du = - csc(x) cot(x) dx .

    Solve that for dx and plug that into the integral. Then you'll easily see what's left over to complete the substitution.
     
  4. Mar 31, 2014 #3
    Thanks. With using your suggestion for a substitution and playing around with the other problem, i have them both figured out.
     
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