Definite Integration Question - None

Question

Statement-1: The value of the integral $\int_{\pi/6}^{\pi/3} \dfrac{dx}{1 + \sqrt{\tan x}}$ is equal to $\dfrac{\pi}{6}$.

Statement-2: $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$.

Accepted Answer

Step 1: Let I = ∫_{π/6}^{π/3} dx/(1 + √(tan x))

Step 2: Apply Statement-2 with a = π/6, b = π/3, so a+b = π/2:

I = ∫_{π/6}^{π/3} 1/(1 + √(tan(π/2 - x))) dx

Step 3: Use tan(π/2 - x) = cot x:

I = ∫_{π/6}^{π/3} 1/(1 + √(cot x)) dx = ∫_{π/6}^{π/3} 1/(1 + 1/√(tan x)) dx

Step 4: Simplify: I = ∫_{π/6}^{π/3} √(tan x)/(√(tan x) + 1) dx

Step 5: Add the original and transformed integral:

2I = ∫_{π/6}^{π/3} [1/(1 + √(tan x)) + √(tan x)/(1 + √(tan x))] dx

2I = ∫_{π/6}^{π/3} [(1 + √(tan x))/(1 + √(tan x))] dx = ∫_{π/6}^{π/3} 1 dx

Step 6: 2I = [x]_{π/6}^{π/3} = π/3 - π/6 = π/6

∴ I = π/12, so Statement-1 is FALSE and Statement-2 is TRUE

Answer: B (Statement-1 is false, Statement-2 is true)

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