## Calculus 8th Edition

$$\int_{\pi/3}^{2\pi/3}\csc^2\frac{t}{2}dt=\frac{4}{\sqrt3}$$
To solve the integral $$\int_{\pi/3}^{2\pi/3}\csc^2\frac{t}{2}dt$$ we will use the substitution $\frac{t}{2}=z$ which gives us $\frac{dt}{2}=dz\Rightarrow dt=2dz$ and the integration bounds would be: for $t=\pi/3$ we have $z=\pi/6$ and for $t=2\pi/3$ we have $z=\pi/3$ so we get: $$\int_{\pi/3}^{2\pi/3}\csc^2\frac{t}{2}dt=\int_{\pi/6}^{\pi/3}csc^2z\cdot2dz=2\left.(-\cot z\right|_{\pi/6}^{\pi/3})=-2(\cot\frac{\pi}{3}-\cot\frac{\pi}{6})=-2(\frac{1}{\sqrt3}-\sqrt3)=-2\frac{1-3}{\sqrt3}=\frac{4}{\sqrt3}$$