Answer
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}cot^{3}x\,dx=\frac{1}{2}(1-ln2)$$
Work Step by Step
$cot^{2}x=csc^{2}x-1,\,{(cotx)}'=-csc^{2}x$
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}cot^{3}x\,dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}cotx(csc^{2}x-1)dx$$
$$=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}cotx\,csc^{2}x\,dx-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{cosx}{sinx}dx$$
$$=-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}cotx\,d(cotx)-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1}{sinx}d(sinx)$$
$$=-\frac{cot^{2}x}{2}-ln(sinx)_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$
$$=\frac{1}{2}(1-ln2)$$
