Using properties of definite integrals, prove that $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\tan x}}=\frac{\pi}{12}$

CBSE Class 12 · 2007 · Definite IntegralsVery Short Answer

Using properties of definite integrals, prove that

\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{d x}{1 + t a n x} = \frac{π}{12}

Answer: Proved

Let

I = \int_{π /6}^{π /3} \frac{d x}{1 + tan x}

. The limits satisfy

a + b = \frac{π}{6} + \frac{π}{3} = \frac{π}{2}

, which is what makes the reflection work.
Apply

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

and use

tan (\frac{π}{2} - x) = cot x

I = \int_{π /6}^{π /3} \frac{d x}{1 + cot x} = \int_{π /6}^{π /3} \frac{tan x}{1 + tan x} d x

Adding the two forms of

I

2 I = \int_{π /6}^{π /3} \frac{1 + tan x}{1 + tan x} d x = \int_{π /6}^{π /3} d x = \frac{π}{3} - \frac{π}{6} = \frac{π}{6}

Hence

I = \frac{π}{12}

. Proved.

CBSE Class 12 Board · 2007 · Subjective