How do you evaluate the integral $int dx/(1-cos3x)$ ?

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Answer 1 · 2017-10-12T23:38:25

$int dx/(1-cos3x) = -1/3cot((3x)/2)+C$

Substitute $t=3x$ :

$int dx/(1-cos3x) = 1/3 int (dt)/(1-cost)$

Use now the trigonometric identity:

$cost= (1-tan^2(t/2))/(1+tan^2(t/2))$

So that:

$1/(1-cost) = 1/(1-(1-tan^2(t/2))/(1+tan^2(t/2)))$

$1/(1-cost) = (1+tan^2(t/2))/((1+tan^2(t/2))-(1-tan^2(t/2)))$

$1/(1-cost) = (1+tan^2(t/2))/(1+tan^2(t/2)-1+tan^2(t/2))$

$1/(1-cost) = (1+tan^2(t/2))/(2tan^2(t/2))$

And as:

$1+tan^2alpha = 1+ sin^2alpha/cos^2alpha = (cos^2alpha+sin^2alpha)/cos^2alpha = 1/cos^2alpha = sec^2alpha$

we get:

$int (dt)/(1-cost) = int sec^2(t/2)/(2tan^2(t/2))dt$

Substitute now:

$u=tan(t/2)$

$du=1/2sec^2(t/2)dt$

and we have:

$int sec^2(t/2)/(2tan^2(t/2))dt = int (du)/u^2 = -1/u+C$

and undoing the substitution:

$int (dt)/(1-cost) = -1/tan(t/2)+C = -cot(t/2)+C$

$int dx/(1-cos3x) = -1/3cot((3x)/2)+C$

How do you evaluate the integral int dx/(1-cos3x)int dx/(1-cos3x)?