# Use NINT to solve the problemEvaluate the integral from 10-0 of 1/(4+3cosx) dx

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You should come up with the trigonometric substitution, such that:

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

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

Changing the limits of integration and variable yields:

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

`int_0^(tan 5) (2dt)(1+t^2))/((t^2+7)(1+t^2))`

Reducing dupliucate terms yields:

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

Using the fundamenatl theorem of calculus yields:

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

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

**Hence, evaluating the given definite integral yields `int_0^10 1/(4+3cos x)dx = 2/sqrt 7 (arctan (tan 5/sqrt 7)).` **