91Ó°ÊÓ

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

a) ${∫}_0^{4\mathrm{π}/3}{\sin }^2\left(\frac{3x}{2}\right)dx$

b)${∫}_{-\mathrm{π}/2}^{3\mathrm{π}/2}{\cos }^2\left(\frac{x}{2}\right)dx$

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

$⇒{∫}_0^{4\mathrm{π}/3}{\sin }^2\left(\frac{3x}{2}\right)dx$a)

Evaluate the following integral without a calculation.

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

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

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

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

Thus, the solution is $I=\frac{2\mathrm{Ï€}}{3}$.

b)

Evaluate the following integral without a calculation.

$⇒{∫}_{-\mathrm{π}/2}^{3\mathrm{π}/2}{\cos }^2\left(\frac{x}{2}\right)dx$

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

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

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

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

Thus, the solution is $I=\mathrm{Ï€}$.