Question

Evaluate the following integral:

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \cot^\frac{3}{2} x}dx\]

 

Answer 1

\[\text{Let I} = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \cot^\frac{3}{2} x}dx..............(1)\]

Then,

\[I = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \cot^\frac{3}{2} \left( \frac{\pi}{3} + \frac{\pi}{6} - x \right)}dx .................\left[ \int_a^b f\left( x \right)dx = \int_a^b f\left( a + b - x \right)dx \right]\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \cot^\frac{3}{2} \left( \frac{\pi}{2} - x \right)}dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \tan^\frac{3}{2} x}dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\cot^\frac{3}{2} x}{\cot^\frac{3}{2} x + 1}dx ...................(2)\]

Adding (1) and (2), we get

\[2I = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1 + \cot^\frac{3}{2} x}{1 + \cot^\frac{3}{2} x}dx\]
\[ \Rightarrow 2I = \int_\frac{\pi}{6}^\frac{\pi}{3} dx\]
\[ \Rightarrow 2I = \left.x\right|_\frac{\pi}{6}^\frac{\pi}{3} \]
\[ \Rightarrow 2I = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}\]
\[ \Rightarrow I = \frac{\pi}{12}\]