Answer
$$\frac{4}{3}$$
Work Step by Step
$$\eqalign{
& \int_{\pi /4}^{\pi /2} {{{\csc }^4}\theta } d\theta \cr
& {\text{We write }}{\csc ^4}\theta {\text{ as }}{\csc ^2}\theta {\csc ^2}\theta \cr
& = \int_{\pi /4}^{\pi /2} {{{\csc }^2}\theta {{\csc }^2}\theta } d\theta \cr
& {\text{use the fundamental identiy }}{\csc ^2}\theta = {\cot ^2}\theta + 1 \cr
& = \int_{\pi /4}^{\pi /2} {\left( {{{\cot }^2}\theta + 1} \right){{\csc }^2}\theta } d\theta \cr
& {\text{Integrate by substitution method}} \cr
& {\text{Let }}u = \cot \theta,\,\,\,\,du = - {\csc ^2}\theta d\theta,\,\,\,\,d\theta = \frac{{du}}{{ - {{\csc }^2}\theta }} \cr
& {\text{Then}}{\text{,}} \cr
& = \int_{}^{} {\left( {{u^2} + 1} \right){{\csc }^2}\theta } \left( {\frac{{du}}{{ - {{\csc }^2}\theta }}} \right) \cr
& {\text{Cancel common factor cs}}{{\text{c}}^2}\theta \cr
& = \int_{}^{} {\left( {{u^2} + 1} \right)} \left( { - du} \right) \cr
& {\text{integrating}} \cr
& = - \frac{1}{3}{u^3} - u + C \cr
& {\text{write in terms of }}\theta .\,\,\,u = \cot \theta \cr
& = - \frac{1}{3}{\left( {\cot \theta } \right)^3} - \cot \theta + C \cr
& \cr
& Then \cr
& \int_{\pi /4}^{\pi /2} {{{\csc }^4}\theta } d\theta = \left( { - \frac{1}{3}{{\left( {\cot \theta } \right)}^3} - \cot \theta } \right)_{\pi /4}^{\pi /2} \cr
& = \left( { - \frac{1}{3}{{\left( {\cot \frac{\pi }{2}} \right)}^3} - \cot \frac{\pi }{2}} \right) - \left( { - \frac{1}{3}{{\left( {\cot \frac{\pi }{4}} \right)}^3} - \cot \frac{\pi }{4}} \right) \cr
& = \left( { - \frac{1}{3}{{\left( 0 \right)}^3} - 0} \right) - \left( { - \frac{1}{3}{{\left( 1 \right)}^3} - 1} \right) \cr
& = \frac{4}{3} \cr} $$
