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#