Use the results ${\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{k}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{k}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\mathit{b}\mathbf{-}\mathit{a}\mathbf{)}$ to evaluate the following integrals without calculation.

(a)${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{4}\mathbf{\pi }\mathbf{/}\mathbf{3}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\left(\frac{3x}{2}\right)\mathit{d}\mathit{x}$

(b)${\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }\mathbf{/}\mathbf{2}}^{\mathbf{3}\mathbf{\pi }\mathbf{/}\mathbf{2}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\left(\frac{x}{2}\right)\mathit{d}\mathit{x}$

${\int }_a^b{\sin }^{2}kxdx={\int }_a^b{\cos }^{2}kxdx=\frac{1}{2}(b-a)$ if $k(b-a)$ is an integral multiples of $\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$. Evaluate the following integral without calculations.

a) ${\int }_0^{4\pi /3}{\sin }^2\left(\frac{3x}{2}\right)dx$

b)${\int }_{-\pi /2}^{3\pi /2}{\cos }^2\left(\frac{x}{2}\right)dx$

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 $\frac{\mathbf{1}}{\mathbf{b}\mathbf{-}\mathbf{a}}{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{d}\mathit{x}$.

$\Rightarrow {\int }_0^{4\pi /3}{\sin }^2\left(\frac{3x}{2}\right)dx$a)

Evaluate the following integral without a calculation.

[Use ${\int }_a^b{\sin }^{2}kxdx={\int }_a^b{\cos }^{2}kxdx=\frac{1}{2}(b-a)$]

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

Substitute the value of a and b in the above equation.

$$\begin{gathered} I=\frac{1}{2}(\frac{4\pi }{3}-0) \\ I=\frac{2\pi }{3} \\ I=\frac{2\pi }{3} \end{gathered}$$

Thus, the solution is $I=\frac{2\pi }{3}$.

b)

Evaluate the following integral without a calculation.

$\Rightarrow {\int }_{-\pi /2}^{3\pi /2}{\cos }^2\left(\frac{x}{2}\right)dx$

[Use ${\int }_a^b{\sin }^{2}kxdx={\int }_a^b{\cos }^{2}kxdx=\frac{1}{2}(b-a)$]

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

Substitute the value of a and b in the above equation.

$$\begin{gathered} I=\frac{1}{2}(\frac{3\pi }{2}-\frac{\left(-\pi \right)}{2}) \\ I=\pi \end{gathered}$$

Thus, the solution is $I=\pi$.