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# 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!

### 1 Answer

- germanoLv 71 decade agoFavorite Answer
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