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Prove that ∫f(x).dx, x ∈ [a, b] = ∫f(a + b - x)dx, x ∈ [a, b] and hence. Evaluate ∫1/1 + √tanx dx, x ∈ [π/3, π/3].

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Best answer

Let t = a + b – x then dt – dx 

When x = a and x = b 

t = b t = a

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