∫ 0 2 π 1 + tan 2 5 4 ( x ) 1 d x
If the above integral evaluates to a number that can be written in the form b a π , where a and b are coprime, positive integers, find a + b .
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Let I be equal to the integral in question and let u = 2 π − x . By Change of Variable, we have I I ⟹ 2 I ⟹ I = ∫ 0 2 π 1 + cot 2 5 4 ( x ) 1 d x = ∫ 0 2 π 1 + tan 2 5 4 ( x ) 1 d x = ∫ 0 2 π 1 + tan 2 5 4 ( x ) 1 + 1 + cot 2 5 4 ( x ) 1 d x = ∫ 0 2 π ( 1 + tan 2 5 4 ( x ) ) ( 1 + cot 2 5 4 ( x ) ) ( 1 + cot 2 5 4 ( x ) ) + ( 1 + tan 2 5 4 ( x ) ) d x = ∫ 0 2 π 2 + cot 2 5 4 ( x ) + tan 2 5 4 ( x ) 2 + cot 2 5 4 ( x ) + tan 2 5 4 ( x ) d x = ∫ 0 2 π 1 d x = 2 π = 4 π Thus, we must have a = 1 and b = 4 , so a + b = 5 .
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Relevant wiki: Integration Tricks
I = ∫ 0 2 π 1 + tan 2 5 4 x 1 d x = 2 1 ∫ 0 2 π 1 + tan 2 5 4 x 1 + 1 + tan 2 5 4 ( 2 π − x ) 1 d x = 2 1 ∫ 0 2 π 1 + tan 2 5 4 x 1 + 1 + cot 2 5 4 x 1 d x = 2 1 ∫ 0 2 π 1 + tan 2 5 4 x 1 + 1 + tan 2 5 4 x 1 1 d x = 2 1 ∫ 0 2 π 1 + tan 2 5 4 x 1 + tan 2 5 4 x + 1 tan 2 5 4 x d x = 2 1 ∫ 0 2 π 1 ⋅ d x = 2 x ∣ ∣ ∣ ∣ 0 2 π = 4 π Using the identity: ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x
⟹ a + b = 1 + 4 = 5