Recent content by everestwitman

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    Help with Trigonometric Integrals

    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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    Help With Arc Length Integral

    How's this? 1. (d/dx)((x^3/6 + 1/(2x)) = x^2/2 - 1/(2x^2) 2. (dy/dx)^2 = (x^4 -1)^2/4x^4 = ((x - 1)^2 (x+1)^2 (x^2+1)^2) / 4x^2 3. (dy/dx)^2 + 1 = (x^4 + 1)^2 / (4x^4) 4. sqrt((dy/dx)^2 + 1) = (x^4+1) / 4x^2 = x^2 / 4 + 1 / (4x^2) 5.∫ x^2 / 4 + 1 / (4x^2) dx from x = 1 to 4 = x^3/12 - 1/...
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    Help with Trigonometric Integrals

    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. Homework Statement 1)∫ 6 csc^3 (x) cot x dx Homework Equations The Attempt at a Solution 6 ∫ csc^3 (x) dx) / tan x csc^3 / tan x =...
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    Help With Arc Length Integral

    Here is the work that i have done so far, corrected for the mistake that you pointed out earlier. 1) Find the length of the curve. y = x^3/6 + 1/(2x) from x = 1 to x = 4 L = ∫sqrt(1 + (d(x^3/6 + 1/(2x)/dx))^2) dx from x = 1 to 4 (d(x^3/6 + 1/(2x)/dx) = (x^3/6 + 1/(2x) ((x^3/6 +...
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    Help With Arc Length Integral

    Now how would I go about factoring the expression that I get when I correct my mistakes? What I have now is ∫sqrt(x^6/36 + x^2/6 + 1/(4x^2) + 1) dx from x = 1 to 4
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    Help With Arc Length Integral

    Thanks a lot for your help, Mark44! I will be sure to in the future.
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    Help With Arc Length Integral

    Here is the work that i have done so far: 1) Find the length of the curve. y = x^3/6 + 1/(2x) from x = 1 to x = 4 L = ∫sqrt(1 + (d(x^3/6 + 1/(2x)/dx))^2) dx from x = 1 to 4 L = ∫sqrt(1 + (x^2/2-1/(2 x^2))^2) dx from x = 1 to 4 L = ∫sqrt(1 + x^4/4 + 1/(x^4) - 1/2) dx from x = 1 to 4 L...
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    Help With Arc Length Integral

    ∫sqrt(x^4/4 + 1/(x^4) + 1/2) dx from x = 1 to 4 Could someone help me solve this? I can't seem to find a substitution that works, or find the square root of (x^4/4 + 1/(x^4). Any help would be very appreciated. Thanks in advance!
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