Question

Evaluate \[ \int_{\frac{\pi}{11}}^{\frac{9\pi}{22}} \frac{dx}{1+\sqrt{\tan x}} \]

• A. \(\dfrac{\pi}{4}\)
• B. \(\dfrac{\pi}{22}\)
• C. \(\dfrac{\pi}{11}\)
• D. \(\dfrac{7\pi}{44}\)

Hint:

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

Answer 1

The Correct Option is D

Step 1: Let the integral be \(I\).
\[ I= \int_{\frac{\pi}{11}}^{\frac{9\pi}{22}} \frac{dx}{1+\sqrt{\tan x}} \] Here, \[ a=\frac{\pi}{11},\quad b=\frac{9\pi}{22} \] Now, \[ a+b=\frac{\pi}{11}+\frac{9\pi}{22} \] \[ a+b=\frac{2\pi}{22}+\frac{9\pi}{22} \] \[ a+b=\frac{11\pi}{22}=\frac{\pi}{2} \]

Step 2: Use the property of definite integrals.
Using the property, \[ \int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx \] we get \[ I= \int_{\frac{\pi}{11}}^{\frac{9\pi}{22}} \frac{dx}{1+\sqrt{\tan\left(\frac{\pi}{2}-x\right)}} \] Since, \[ \tan\left(\frac{\pi}{2}-x\right)=\cot x \] therefore, \[ I= \int_{\frac{\pi}{11}}^{\frac{9\pi}{22}} \frac{dx}{1+\sqrt{\cot x}} \]

Step 3: Add the two forms of the integral.
Now, \[ 2I= \int_{\frac{\pi}{11}}^{\frac{9\pi}{22}} \left[ \frac{1}{1+\sqrt{\tan x}} + \frac{1}{1+\sqrt{\cot x}} \right]dx \] Let \[ \sqrt{\tan x}=t \] Then, \[ \sqrt{\cot x}=\frac{1}{t} \] So, \[ \frac{1}{1+t}+\frac{1}{1+\frac{1}{t}} = \frac{1}{1+t}+\frac{t}{1+t} \] \[ = \frac{1+t}{1+t}=1 \] Therefore, \[ 2I= \int_{\frac{\pi}{11}}^{\frac{9\pi}{22}}1\,dx \]

Step 4: Evaluate the definite integral.
\[ 2I= \frac{9\pi}{22}-\frac{\pi}{11} \] \[ 2I= \frac{9\pi}{22}-\frac{2\pi}{22} \] \[ 2I=\frac{7\pi}{22} \] Hence, \[ I=\frac{7\pi}{44} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{7\pi}{44}} \]