Answer
a. $F(x)=-cot(x)+C$
b. $F(x)=cot(\frac{3x}{2})+C$
c. $F(x)=x+4cot(2x)+C$
Work Step by Step
a. Given $f(x)=csc^2x$, we have $F(x)=-cot(x)+C$ as its antiderivative, where $C$ is a constant.
b. Given $f(x)=-\frac{3}{2}csc^2(\frac{3x}{2})$, we have $F(x)=cot(\frac{3x}{2})+C$ as its antiderivative, where $C$ is a constant.
c. Given $f(x)=1-8csc^2(2x)$, we have $F(x)=x+4cot(2x)+C$ as its antiderivative, where $C$ is a constant.