Answer
$$\int_{\pi/4}^{\pi/2}cot^{3}x\,dx=\frac{1-ln2}{2}$$
Work Step by Step
$$\int_{\pi/4}^{\pi/2}cot^{3}x\,dx=\int_{\pi/4}^{\pi/2}cot\,x(csc^{2}x-1)dx$$
$$=\int_{\pi/4}^{\pi/2}(cot\,x\,csc^{2}x-cot\,x)dx$$
$$=-\int_{\pi/4}^{\pi/2}cot\,x\,d(cot\,x)-\int_{\pi/4}^{\pi/2}\frac{cos\,x}{sin\,x}\,dx$$
$$=\left [ -\frac{1}{2}cot^{2}x-ln(sin\,x) \right ]_{\pi/4}^{\pi/2}$$
$$=\frac{1-ln2}{2}$$
