Evaluate the integral #int \ x/(cosx-1) \ dx #?
1 Answer
# int \ x/(cosx-1) \ dx = xcot(x/2) - 2ln|sin(x/2)| + C #
Explanation:
We seek:
# I = int \ x/(cosx-1) \ dx # ..... [A]
We can apply Integration By Parts, but prior to this, let us consider:
# I_1 = int \ 1/(cosx-1) \ dx #
For this we can use the tangent half angle identity:
# cos alpha = (1-tan^2(alpha/2))/(1+tan^2(alpha/2)) #
Which gives us:
# I_1 = int \ 1/((1-tan^2(x/2))/(1+tan^2(x/2))-1) \ dx #
# \ \ \ = int \ 1/((1-tan^2(x/2)- (1+tan^2(x/2)))/(1+tan^2(x/2))) \ dx #
# \ \ \ = int \ (1+tan^2(x/2))/( 1-tan^2(x/2) - 1-tan^2(x/2) ) \ dx #
# \ \ \ = int \ (1+tan^2(x/2))/( -2tan^2(x/2) ) \ dx #
# \ \ \ = int \ (sec^2(x/2))/( -2tan^2(x/2) ) \ dx #
We can perform a substitution of the form:
# u=tan(x/2) => (du)/dx = 1/2sec^2(x/2) #
If we substitute into the integral, we get:
# I_1 = int -1/u^2 \ du #
# \ \ \ = 1/u + C #
And restoring the earlier substitution, we get:
# I_1 = 1/tan(x/2) + C #
# \ \ \ = cot(x/2) + C #
Now, let us return to the original integral [A], and apply Integration By Parts, using this result:
Let
# { (u,=x, => (du)/dx,=1), ((dv)/dx,=1/(cosx-1), => v,=cot(x/2) ) :}#
Then plugging into the IBP formula:
# int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx #
Gives us
# int \ (x)(1/(cosx-1)) \ dx = (x)(cot(x/2)) - int \ (cot(x/2))(1) \ dx #
# :. I = xcot(x/2) - int \ cot(x/2) \ dx + C #
And now we can manipulate the second integral:
# I_2 = int \ cot(x/2) \ dx #
# \ \ \ = int \ cos(x/2)/sin(x/2) \ dx #
We can apply the substitution
# u = sin(x/2) => (du)/dx = 1/2cos(x/2) #
Substituting, we get:
# I_2 = int \ (2)(1/2cos(x/2))/sin(x/2) \ dx #
# \ \ \ = int \ (2)(1/u) \ du #
# \ \ \ = 2ln|u| #
And restoring the substitution we get:
# I_2 = 2ln|sin(x/2)| #
Then, combining our results we have:
# I = xcot(x/2) - 2ln|sin(x/2)| + C #