## Thomas' Calculus 13th Edition

$$\frac{4}{3}$$
\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}