Question

\(\int\limits_\frac{\pi}{6}^\frac{\pi}{3}\frac{1}{1+\sqrt{cotx}}dx=\)

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

Answer 1

The Correct Option is A

To solve the integral \(\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\sqrt{\cot x}}\, dx\), we first proceed by making a substitution. Consider the substitution \( x = \frac{\pi}{2} - u \), meaning \( dx = -du \). Under this substitution, cotangent changes as follows: \(\cot x = \tan u\). The limits will also change: When \( x = \frac{\pi}{6} \), \( u = \frac{\pi}{3} \), and when \( x = \frac{\pi}{3} \), \( u = \frac{\pi}{6} \). Now, rewrite the integral:

\(-\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{6}}\frac{1}{1+\sqrt{\tan u}}\, du\)

Reversing the limits changes the sign:

\(\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\sqrt{\tan u}}\, du\)

Observing that the transformed integral is identical to the original, it implies the value of the integral is half of the full measure between these limits. Given the symmetrical properties of the function, we can equate:

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

Solving the right-hand side:

\( = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \)

Thus, \( I = \frac{\pi}{12} \).

The value of the integral is \(\frac{\pi}{12}\).