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    The integrals of (tanh2x)^3:

    1) (1/2)ln(cosh2x) - (1/4)(tanh2x)^2 +c

    2) (1/2)ln(cosh2x) + (1/4)(sech2x)^2 +c

    Since in the part that needs integrating there is: (sech^2(2x))tanh2x)

    So I thought I could do: let y=sech^2(2x)

    then dy/dx = 4sech2x(-sech2xtanh2x) = -4sech^2(2x)tanh2x

    so I = -1/4(sech^2(2x))

    and then I include this overall and get (2)

    What do you think?


    Thank you!

    I wanted to post this in the maths forum.
    • PS Helper
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    Well 1 - \tanh^2 u = \text{sech}^2\, u, so they're actually the same answer; the difference is accounted for in the arbitrary constant of integration.
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    Let's imagine an integral does have two different solutions.

    If you differentiate their difference... what do you get?
    And what does that tell you?
    • Study Helper
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    Here's the two functions in red and green (assuming the constant is zero), and finally in brown at the bottom is the difference between the two.
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    • Thread Starter
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    Thank you all
 
 
 
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