Question

\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]

Answer 1

`int 1/[ 2 - 3cos2x] dx`

As `cos 2x = 2cos^x - 1`

So `int 1/[ 2 - 3cos2x] dx = int 1/[2 - 3( 2cos^x - 1) ]`

And multiply and divide by sec²x
Then we have `int  sec^2x/[5sec^2x - 6]` dx

= `int   (sec^2x)/[5( 1 + tan^2x) - 6]`

=  `int (sec^2x dx)/( 5tan^2x - 1)`

Let tan x = t, then sec²x dx = dt

Hence `int [sec^2x]/[ 5tan^2x - 1]` 

= `int dt/[5t^2 - 1]`

= `1/5 int  dt/[t^2 - (1/sqrt5)^2]`

= `1/5 log |[ t - 1/sqrt5 ]/[ t + 1/sqrt5 ]|`

= `1/5 log |[ tan x - 1/sqrt5]/[ tan x + 1/sqrt5 ]| + c`