_{π/6}^{π/3} {1}{1+{ x}} d x=

Correct Answer is : π12π/1212π

Correct Answer is : π12π/1212π

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COMEDK 2025 Evening Solved Paper

Question : 168 of 180

Marks: +1, -0

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

Solution:

Concept:

Use the property

\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx

with

a=\frac{\pi}{6}, b=\frac{\pi}{3}

to simplify the integrand.

Explanation:

Let

I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan x}}\,dx

Apply the substitution

x \to \frac{\pi}{2} - x

(since

\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}

Then

dx

changes sign, but reversing the limits keeps the integral same, so we have:

I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan(\frac{\pi}{2}-x)}}\,dx

Since

\tan(\frac{\pi}{2}-x) = \cot x = \frac{1}{\tan x}

, we get

\sqrt{\tan(\frac{\pi}{2}-x)} = \frac{1}{\sqrt{\tan x}}

Thus the integrand becomes

\frac{1}{1+\frac{1}{\sqrt{\tan x}}} = \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}

Adding the two expressions for

I

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

Compute the integral:

\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1\,dx = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}

Therefore,

2I = \frac{\pi}{6}

, so

I = \frac{\pi}{12}

Answer:

Option C,

\frac{\pi}{12}

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