Question

\( \int_{\pi/11}^{9\pi/22} \frac{dx}{1 + \sqrt{\tan x}} \) is equal to:

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

Hint:

When dealing with definite integrals and symmetric limits, use substitution and properties of even and odd functions to simplify the expression.

Answer 1

The Correct Option is D

we have given integral is: \[ I = \int_{\pi/11}^{9\pi/22} \frac{dx}{1 + \sqrt{\tan x}}. \] To simplifying the integral, rewrite it as: \[ I = \int_{\pi/11}^{9\pi/22} \frac{\sqrt{\cos x} \, dx}{\sqrt{\sin x} + \sqrt{\cos x}}. \] Next, perform a substitution: Let \[ x \to \frac{\pi}{2} - x. \] Under this substitution, the following transformations occur: \[ \sin x \to \cos x, \quad \cos x \to \sin x, \quad dx \to -dx. \] This changes the integral to: \[ I = \int_{\pi/11}^{9\pi/22} \frac{\sqrt{\sin x} \, dx}{\sqrt{\cos x} + \sqrt{\sin x}}. \] Let this new integral be denoted as \( I_2 \). Now, by adding the original integral \( I \) and the substituted integral \( I_2 \), we get: \[ I + I_2 = \int_{\pi/11}^{9\pi/22} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx. \] Simplifying the right-hand side: \[ I + I_2 = \int_{\pi/11}^{9\pi/22} 1 \, dx. \] The limits of integration give: \[ \int_{\pi/11}^{9\pi/22} 1 \, dx = \left[ x \right]_{\pi/11}^{9\pi/22} = \frac{9\pi}{22} - \frac{\pi}{11}. \] Simplifying this result: \[ \frac{9\pi}{22} - \frac{\pi}{11} = \frac{9\pi}{22} - \frac{2\pi}{22} = \frac{7\pi}{22}. \] Thus, \( I + I_2 = \frac{7\pi}{22} \). Since \( I = I_2 \), we have: \[ 2I = \frac{7\pi}{22}. \] Solving for \( I \): \[ I = \frac{7\pi}{44}. \] Final Answer: \[ \boxed{\frac{7\pi}{44}} \]