Đáp án:
$\begin{array}{l}
\left( {\cot x} \right)' = - \frac{1}{{{{\sin }^2}x}}\\
\frac{1}{{{{\sin }^2}x}} = {\cot ^2}x + 1\\
\Rightarrow \int\limits_{\pi /4}^{\pi /2} {\frac{{\sqrt {\cot x} }}{{{{\sin }^4}x}}dx} \\
= \int\limits_{\pi /4}^{\pi /2} {\frac{{\sqrt {\cot x} }}{{{{\sin }^2}x}}\left( { - 1} \right).d\left( {\cot x} \right)} \\
= - \int\limits_{\pi /4}^{\pi /2} {\sqrt {\cot x} .\left( {{{\cot }^2}x + 1} \right)d\left( {\cot x} \right)} \\
= - \int\limits_{\pi /4}^{\pi /2} {{{\cot }^{\frac{5}{2}}}x + {{\cot }^{\frac{1}{2}}}xd\cot x} \\
= - \left( {\frac{2}{7}.{{\cot }^{\frac{7}{2}}}x + \frac{2}{3}.{{\cot }^{\frac{3}{2}}}x} \right)_{\pi /4}^{\pi /2}\\
= - \frac{{20}}{{21}}
\end{array}$
