Question

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

• A. \( \frac{\pi}{3} \)
• B. \( \frac{\pi}{6} \)
• C. \( \frac{\pi}{12} \)
• D. \( \frac{\pi}{2} \)

Hint:

For definite integrals with limits whose sum is \( \frac{\pi}{2} \), use \( x \to \frac{\pi}{2}-x \)This often converts \( \tan x \) into \( \cot x \) and simplifies the integral.

Answer 1

The Correct Option is C

Step 1: Let the integral be \( I \).
\[ I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan x}}\,dx \]

Step 2: Use the definite integral property.
\[ \int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx \]
Here:
\[ a=\frac{\pi}{6},\quad b=\frac{\pi}{3} \]
So:
\[ a+b=\frac{\pi}{2} \]

Step 3: Apply the property.
\[ I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan\left(\frac{\pi}{2}-x\right)}}\,dx \]

Step 4: Simplify using trigonometric identity.
\[ \tan\left(\frac{\pi}{2}-x\right)=\cot x=\frac{1}{\tan x} \]
So:
\[ \sqrt{\tan\left(\frac{\pi}{2}-x\right)} = \frac{1}{\sqrt{\tan x}} \]

Step 5: Rewrite the transformed integral.
\[ I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\frac{1}{\sqrt{\tan x}}}\,dx \]
\[ I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}\,dx \]

Step 6: Add both forms of \( I \).
\[ 2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left[ \frac{1}{1+\sqrt{\tan x}} + \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} \right]dx \]
\[ 2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1\,dx \]

Step 7: Final conclusion.
\[ 2I=\frac{\pi}{3}-\frac{\pi}{6} \]
\[ 2I=\frac{\pi}{6} \]
\[ I=\frac{\pi}{12} \]
\[ \boxed{\frac{\pi}{12}} \]