Question

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]

Answer 1

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]
\[ = \int_0^\frac{\pi}{4} \frac{1}{\left( \tan x + \cot x \right)^2}dx\]
\[ = \int_0^\frac{\pi}{4} \frac{1}{\left( \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \right)^2}dx\]
\[ = \int_0^\frac{\pi}{4} \frac{1}{\left( \frac{\sin^2 x + \cos^2 x}{\sin x\cos x} \right)^2}dx\]
\[ = \int_0^\frac{\pi}{4} \sin^2 x \cos^2 xdx\]

\[= \frac{1}{4} \int_0^\frac{\pi}{4} \left( 2\sin x\cos x \right)^2 dx\]
\[ = \frac{1}{4} \int_0^\frac{\pi}{4} \sin^2 2xdx\]
\[ = \frac{1}{4} \int_0^\frac{\pi}{4} \left( \frac{1 - \cos4x}{2} \right)dx\]
\[ = \frac{1}{8} \int_0^\frac{\pi}{4} dx - \frac{1}{8} \int_0^\frac{\pi}{4} \cos4xdx\]
\[ = \left.\frac{1}{8} x\right|_0^\frac{\pi}{4} - \left.\frac{1}{8} \left( \frac{\sin4x}{4} \right)\right|_0^\frac{\pi}{4}\]

\[= \frac{1}{8}\left( \frac{\pi}{4} - 0 \right) - \frac{1}{32}\left(\sin \pi - \sin0 \right)\]
\[ = \frac{\pi}{32} - \frac{1}{32} \times \left( 0 - 0 \right)\]
\[ = \frac{\pi}{32}\]