# Evaluate `int_0^10((1)/(4+3cosx))dx`

You need to use the following trigonometric substitution, such that:

`tan (x/2) = t => (1 + tan^(x/2))/2 dx = dt => dx = (2dt)/(1 + t^2)`

You need to write `cos x` in terms of `tan(x/2)` , such that:

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

Changing the limits of integration, yields:

`x = 0 => tan 0 = 0 => t = 0`

`x = 10 => tan 5 = t `

Changing the variable, yields:

`int_0^(tan 5) 1/(4 + 3(1 - t^2)/(1 + t^2))((2dt)/(1 + t^2))`  = `2*int_0^(tan 5) (1 + t^2)/(4 + 4t^2 + 3 - 3t^2)*(dt)/(1 + t^2)`

Reducing duplicate factors yields:

`2 int_0^(tan 5) (dt)/(t^2 + 7) = (2/sqrt7) (arctan (t/sqrt7))|_0^(tan 5)`

Using the fundamental theorem of calculus, yields:

`2 int_0^(tan 5) (dt)/(t^2 + 7) = (2/sqrt7)(arctan ((tan5)/sqrt7) - arctan 0)`

`2 int_0^(tan 5) (dt)/(t^2 + 7) = (2/sqrt7)arctan ((tan5)/sqrt7`

Hence, evaluating the given definite integral, using trigonometric substitution, yields `int_0^10 1/(4 + 3cos x) dx = (2/sqrt7)arctan ((tan5)/sqrt7.`

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