Q75 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics

$\int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \dfrac{1}{1+\sqrt{\tan x}} d x=$

  • A
    $\dfrac{\pi}{3}$
  • B
    $\dfrac{\pi}{2}$
  • C
    $\dfrac{\pi}{6}$
  • Correct answer
    D
    $\dfrac{\pi}{12}$

Solution

Let $I = \int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \dfrac{1}{1 + \sqrt{\tan x}} dx$.

Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$, we have $a + b = \dfrac{\pi}{6} + \dfrac{\pi}{3} = \dfrac{\pi}{2}$.

$I = \int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \dfrac{1}{1 + \sqrt{\tan(\dfrac{\pi}{2} - x)}} dx = \int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \dfrac{1}{1 + \sqrt{\cot x}} dx$.

$I = \int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \dfrac{1}{1 + \dfrac{1}{\sqrt{\tan x}}} dx = \int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \dfrac{\sqrt{\tan x}}{\sqrt{\tan x} + 1} dx$.

Adding the two expressions for $I$:

$2I = \int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \left( \dfrac{1}{1 + \sqrt{\tan x}} + \dfrac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} \right) dx = \int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} 1 dx$.

$2I = [x]_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} = \dfrac{\pi}{3} - \dfrac{\pi}{6} = \dfrac{\pi}{6}$.

$I = \dfrac{\pi}{12}$.

Answer: $\dfrac{\pi}{12}$

Hence, the correct answer is option: 'D'.

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