The equation above holds true for coprime positive integers. Find .
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Divide top and bottom of the integrand by 4 cos 2 ( x ) to end up with
4 1 ∫ 0 2 π 1 + ( 2 3 tan ( x ) ) 2 sec 2 ( x ) d x .
Now let u = 2 3 tan ( x ) . Then d u = 2 3 sec 2 ( x ) d x , transforming our integral to
6 1 ∫ 1 + u 2 1 d u = 6 1 tan − 1 ( u ) = 6 1 tan − 1 ( 2 3 tan ( x ) ) ,
which we must then evaluate from x = 0 to x = 2 π , (the latter approached from the left). This yields the value
6 1 x → 2 π − lim tan − 1 ( 2 3 tan ( x ) ) − 6 1 tan − 1 ( 2 3 tan ( 0 ) ) = 6 1 × 2 π − 0 = 1 2 π ,
giving us A + B = 1 + 1 2 = 1 3 .