Consider the integral `int_0^(2pi)(dx)/(5-2cosx)` making the substitution `tan1/x=t ,` we have `I=int_0^(2pi)(dx)/(5-2cosx)` `=int_0^0(2dt)/((1+t^2)[(5-2(1-t^2))/((1+t^2))])=0` The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.

Answer

2

Explanation

Here the mistake lies in the substitution `"tan"1/2x=t`, because `"tan"1/2x` is discontinuous at `x=pi` which is a point in the interval `[0,2pi]`.