Statement‑I : The value of the integral \int_{\pi / 6}^{\pi / 3} \frac{dx}{1 + \sqrt{tanx}} is equal to \frac{\pi}{6}. Statement‑II : \int_{a}^{b} f \left(\right. x \left.\right) \textrm{ } dx = \int_{a}^{b} f \left(\right. a + b

Engineering

Mathematics

Properties of Definite Integral

Question

Statement‑I : The value of the integral $\int_{π / 6}^{π / 3} \frac{dx}{1 + \sqrt{tanx}}$ is equal to $\frac{π}{6}$ .

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

Statement‑I is true, Statement‑II is true, Statement‑II is not a correct explanation for Statement‑I.

Statement‑I is true, Statement‑II is true, Statement‑II is a correct explanation for Statement‑I.

Statement‑I is false, Statement‑II is true.

Statement‑I is true, Statement‑II is false.

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Let $I = \int_{π / 6}^{π / 3} \frac{\sqrt{cosx}}{\sqrt{cosx} + \sqrt{sinx}} dx$

By using property $\int_{a}^{b} f (x) dx = \int_{a}^{b} f (a + b - x) dx$

$I = \int_{π / 6}^{π / 3} \frac{\sqrt{sinx}}{\sqrt{cosx} + \sqrt{sinx}} dx$

On adding above two equations, we get

$2 I = \int_{π / 6}^{π / 3} \frac{\sqrt{cosx} + \sqrt{sinx}}{\sqrt{cosx} + \sqrt{sinx}} dx = {[x]}_{π / 6}^{π / 3}$

$2 I = \frac{π}{3} - \frac{π}{6} = \frac{π}{6}$

$I = \frac{π}{12}$

Statement‑I is false, Statement‑II is correct.

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