Question

Evaluate the integral `int 1/(5+3cosx) dx`?

Answer 1

`int \ 1/(5+3cosx) \ dx = 1/2 arctan(1/2 tan (x/2)) +C`

Explanation:

We want to evaluate:

`I = int \ 1/(5+3cosx) \ dx`

If we use the trigonometry half angle tangent formula then we have:

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

Let us perform the requested change of variable via the substation:

`tan (x/2) = u`

Differentiating wrt `x` and using the identity `1+tan^2 alpha = sec^2 alpha` we get:

`(du)/dx = 1/2 sec^2(u/2)`
`" " = 1/2 (1+tan^2(u/2))`
`" " = 1/2 (1+u^2)`

`:. 2/(1+u^2) \ (du)/dx = 1`

Then substituting into the integral, we get:

`I = int \ 1/(5+3 \ (1-u^2) / (1+u^2)) \ 2/(1+u^2) \ du`

`\ \ = int \ 1/((5(1+u^2)+3(1-u^2))/(1+u^2)) \ 2/(1+u^2) \ du`

`\ \ = int \ (1+u^2)/(5+5u^2+3-3u^2) \ 2/(1+u^2) \ du`

`\ \ = int \ 2/(8+2u^2) \ du`

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

The above is a standard integral than we can quote

`I = 1/2 arctan(u/2) +C`

Restoring the substitution we get:

`I = 1/2 arctan((tan (x/2))/2) +C`
`\ \ = 1/2 arctan(1/2 tan (x/2)) +C`