Question

Calculate the following integral:
${\int }_{\frac{\pi }{8}}^{\frac{\pi }{4}}{\cot }^{2}2xdx$

Answer 1

${\int }_{\frac{\pi }{8}}^{\frac{\pi }{4}}{\cot }^2\left(2x\right)dx={\int }_{\frac{\pi }{8}}^{\frac{\pi }{4}}-1+{\csc }^2\left(2x\right)dx=-{\int }_{\frac{\pi }{8}}^{\frac{\pi }{4}}dx+{\int }_{\frac{\pi }{8}}^{\frac{\pi }{4}}{\csc }^2\left(2x\right)dx$

Apply substitution $u=2x\frac{1}{2}du=dx$

$=-[\frac{\pi }{4}-\frac{\pi }{8}]+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}cs{c}^2\left(u\right)\left(\frac{1}{2}du\right)$

$=\frac{-\pi }{8}+\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}cs{c}^2\left(u\right)du$

$=\frac{-\pi }{8}+\frac{1}{2}[-\cot u{|}_{\frac{\pi }{4}}^{\frac{\pi }{2}}]$

$=\frac{-\pi }{8}+\frac{1}{2}[-\cot \frac{\pi }{2}+\cot \frac{\pi }{4}]$

$=\frac{1}{2}-\frac{\pi }{8}$

$=\left(\frac{4-\pi }{8}\right)$.