Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 30

$$\int\csc\Big(\frac{v-\pi}{2}\Big)\cot\Big(\frac{v-\pi}{2}\Big)dv=-2\csc\Big(\frac{v-\pi}{2}\Big)+C$$

$$A=\int\csc\Big(\frac{v-\pi}{2}\Big)\cot\Big(\frac{v-\pi}{2}\Big)dv$$ We set $u=\csc\Big(\frac{v-\pi}{2}\Big)$ Then $$du=-\Big(\frac{v-\pi}{2}\Big)'\csc\Big(\frac{v-\pi}{2}\Big)\cot\Big(\frac{v-\pi}{2}\Big)dv$$ $$du=-\Big(\frac{1}{2}-0\Big)\csc\Big(\frac{v-\pi}{2}\Big)\cot\Big(\frac{v-\pi}{2}\Big)dv$$ $$du=-\frac{1}{2}\csc\Big(\frac{v-\pi}{2}\Big)\cot\Big(\frac{v-\pi}{2}\Big)dv$$ That means $$\csc\Big(\frac{v-\pi}{2}\Big)\cot\Big(\frac{v-\pi}{2}\Big)dv=-2du$$ Therefore, $$A=\int-2du=-2u+C$$ $$A=-2\csc\Big(\frac{v-\pi}{2}\Big)+C$$