Question

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

Answer 1

`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`