Evaluate the integral #int \ x/(cosx-1) \ dx #?

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 #