`int 1/(cos(theta) -1) d theta` Find the indefinite integral

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Explanation for `cos(x) = (1-u^2)/(1+u^2)`

before that , we know

`cos(2x)= cos^2(x) -sin^2(x)`

as `cos^2(x)` can be written as `1/(sec^2(x))`

and we can show `sin^2(x) = ((sin^2(x))/(cos^2(x) ))/(1/(cos^2(x)))`

 = `tan^2(x)/sec^2x`

so now ,

`cos(2x)= cos^2(x) -sin^2(x)`

= `(1/sec^2(x)) - (tan^2(x)/sec^2(x))`

=`(1-tan^2(x))/(sec^2(x))`

but `sec^2(x) = 1+tan^2(x)` ,as its an identity

so,

=`(1-tan^2(x))/(sec^2(x))`

=`(1-tan^2(x))/(1+(tan^2(x)))`

so ,

`cos(2x) = (1-tan^2(x))/(1+(tan^2(x)))`

so,

then

`cos(x) = (1-tan^2(x/2))/(1+(tan^2(x/2)))`

as before we told to assume that `u= tan(x/2),`

so,

`cos(x) = (1-u^2)/(1+u^2)`

Hope this helps to understand better

Approved by eNotes Editorial
An illustration of the letter 'A' in a speech bubbles

Given to solve

`int 1/(cos(theta) - 1) d theta`

For convenience, let  `theta = x`

=>

`int 1/(cosx - 1) dx`

let `u = tan(x/2) => dx = (2/(1+u^2)) du`

so ,

`cos(x) = (1-u^2)/(1+u^2)` (See my reply below for an explanation)

so,

`int 1/(cos(x) - 1) dx`

= `int 1/((1-u^2)/(1+u^2) - 1) (2/(1+u^2)) du`

= `int 1/(((1-u^2)-(1+u^2))/(1+u^2) ) (2/(1+u^2)) du`

=`int (1+u^2)/(((1-u^2)-(1+u^2)) ) (2/(1+u^2)) du`

=`int (2)/(((1-u^2)-(1+u^2)) ) du`

=`int (2)/(((1-u^2)-1-u^2)) ) du`

= `int (2)/(-2u^2) du`

=` -int(1/u^2) du`

= `-[u^(-2+1)/(-2+1)]`

= `u^-1`

= `1/u`

= `1/tan(x/2) `

= `cot(x/2)+c`

But `x= theta`

so,

`int 1/(cos(theta) - 1) d theta = cot(theta/2)+c`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial