I=∫π6π311+cotx dx........(1) Apply ∫abf(x) dx=∫abf(a+b−x) dx I=∫π6π311+tanx dx=∫π6π3cotxcotx+1 dx........(2) (1)+(2) ,2I=∫π6π31+cotx1+cotx dx= [x]π6π3=π3−π6=π6 ∴I=π12.

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$\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \sqrt{\cot x}} d x$

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$\frac{π}{4}$

$\frac{π}{12}$

$\frac{π}{6}$

$\frac{π}{2}$

answer is C.

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Detailed Solution

$I = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \sqrt{\cot x}} d x ........ (1) A p p l y \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x I = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \sqrt{\tan x}} d x = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + 1} d x ........ (2) (1) + (2), 2 I = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1 + \sqrt{\cot x}}{1 + \sqrt{\cot x}} d x = {[x]}_{\frac{π}{6}}^{\frac{π}{3}} = \frac{π}{3} - \frac{π}{6} = \frac{π}{6} ∴ I = \frac{π}{12} .$