Chapter 7.4, Problem 13P

Using (4.3) and equations similar to (4.5) to (4.7), show that ${\displaystyle {\int }_a^b{\sin }^2}kxdx={\displaystyle {\int }_a^b{\cos }^2}kxdx=\frac{1}{2}(b-a)$ if $k(b-a)$ is an integral multiple of $\pi ,$ or if kb and ka are both integral multiples of $\pi /2$.