Integral of sec(3x)/tan(3x) please help?

same as integral 1/(tan(3x)sec(3x) or cot(3x)sec(3x)

does anyone know how to do this? Thanks!

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  • 1 decade ago
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    Hello,

    ∫ [sec(3x) /tan(3x)] dx =

    let's rewrite it as:

    ∫ sec(3x) [1/tan(3x)] dx =

    ∫ sec(3x) cot(3x) dx =

    where sec(3x) cot(3x) = [1/cos(3x)][cos(3x)/sin(3x)] = [1 /sin(3x)] = csc(3x):

    ∫ csc(3x) dx =

    multiply the integrand by (1/3) {3[- cot(3x) + csc(3x)] /[csc(3x) - cot(3x)]} (= 1):

    ∫ (1/3) 3[- cot(3x) + csc(3x)] csc(3x) dx /[csc(3x) - cot(3x)] =

    (pulling out 1/3)

    (1/3) ∫ 3[- cot(3x) + csc(3x)] csc(3x) dx /[csc(3x) - cot(3x)] =

    expand the top:

    (1/3) ∫ [- 3cot(3x) + 3csc(3x)] csc(3x) dx /[csc(3x) - cot(3x)] =

    (1/3) ∫ [- 3cot(3x) csc(3x) + 3csc²(3x)] dx /[csc(3x) - cot(3x)] =

    note that the top is the derivative of he bottom:

    (1/3) ∫ d[csc(3x) - cot(3x)] /[csc(3x) - cot(3x)] =

    ending up with:

    (1/3) ln |csc(3x) - cot(3x)| + C

    in conclusion:

    ∫ [sec(3x) /tan(3x)] dx = (1/3) ln |csc(3x) - cot(3x)| + C

    I hope it helps

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