Show that ${\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{k}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{k}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\mathit{b}\mathbf{-}\mathit{a}\mathbf{)}$ if$\mathit{k}\mathbf{(}\mathit{b}\mathbf{}\mathbf{-}\mathbf{}\mathit{a}\mathbf{)}$is an integral multiple of$\mathit{\pi }$, or if kb and ka are both integral multiples of $\frac{\mathbf{\pi }}{\mathbf{2}}$.

${\int }_a^b{\sin }^{2}kxdx={\int }_a^b{\cos }^{2}kxdx=\frac{1}{2}(b-a)$If $k(b-a)$ is an integral multiples of $\pi /2$.

The average value of a function over a particular interval can be found with an expression involving an integral.

Let's say that interval is $\mathbf{[}\mathit{a}\mathbf{,}\mathit{b}\mathbf{|}$.

Then the average value of f(x) over said interval is $\frac{\mathbf{1}}{\mathbf{b}\mathbf{-}\mathbf{a}}{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{d}\mathit{x}$.

To show that:

${\int }_a^b{\sin }^{2}kxdx={\int }_a^b{\cos }^{2}kxdx=\frac{1}{2}(b-a)$ if $k(b-a)$is an integral multiple of $\pi /2$.

${\int }_a^b{\sin }^{2}kxdx+{\int }_a^b{\cos }^{2}dx={\int }_a^{b}dx=b-a$ ...... (1)

With the help of average value method calculate as shown below.

$$\begin{gathered} {\int }_a^b{\sin }^2\left(kx\right)dx=\frac{1}{k}{\int }_{ka}^{kb}{\sin }^{2}tdt \\ =\frac{1}{2k}{\int }_{ak}^{bk}(1-\cos (2t\left)\right)dt \\ =\frac{1}{2k}\left[k(b-a)-{\left.\frac{1}{2}\sin \left(2t\right)\right|}_{ak}^{bk}\right] \end{gathered}$$

$=\frac{1}{2}(b-a)$ ...... (2)

The second equation disappears since$ka=n\pi /2$ and$kb=n\pi /2,n\in \mathrm{\mathbb{Z}}$.

Use equation first and second:

${\int }_a^b{\cos }^{2}dx=\frac{1}{2}(b-a)$

Thus, the solution is ${\int }_a^b{\sin }^{2}kxdx={\int }_a^b{\cos }^{2}kxdx=\frac{1}{2}(b-a)$.