Question

(a) Prove that ${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\pi }\mathbf{/}\mathbf{2}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\pi }\mathbf{/}\mathbf{2}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{x}\mathit{d}\mathit{x}$by making the change of variable$\mathit{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{}\mathit{\pi }\mathbf{-}\mathit{t}$ in one of the integrals.

(b) Use the same method to prove that the averages of$\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{(}\mathit{n}\mathit{\pi }\mathit{x}\mathbf{/}\mathit{l}\mathbf{)}$ and$\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{(}\mathit{n}\mathit{\pi }\mathit{x}\mathbf{/}\mathit{l}\mathbf{)}$ are the same over a period.

Short answer

(a).The solution is ${\int }_0^{\mathrm{\pi }/2}si{n}^{2}xdx={\int }_0^{\mathrm{\pi }/2}co{s}^{2}xdx$.

(b). The solution is .${\int }_0^{\mathrm{\pi ll}2}si{n}^2\frac{n\pi x}{l}dx={\int }_0^{\mathrm{\pi ll}2}co{s}^2\frac{n\pi x}{l}dx$

Step-by-step solution

Given information

Prove that ${\int }_0^{\mathrm{\pi }/2}si{n}^{2}xdx={\int }_0^{\mathrm{\pi }/2}co{s}^{2}xdx$ by making the change of variable $x=\frac{1}{2}\pi -t$in one of the integral.

Solve the integral from given information

a).

From the given information, ${\int }_0^{\mathrm{\pi }/2}si{n}^{2}xdx={\int }_0^{\mathrm{\pi }/2}co{s}^{2}xdx$ by making the change of variable $x=\frac{1}{2}\pi -t$ in one of the integrals is:

$$\begin{gathered} d{\int }_0^{\mathrm{\pi }/2}si{n}^{2}xdx={\int }_{\mathrm{\pi }/2}^0{\sin }^2\left(\frac{\pi }{2}-t\right)\left(-dt\right) \\ ={\int }_{\mathrm{\pi }/2}^{\pi }{\cos }^{2}tdt \end{gathered}$$

Thus, the solution is ${\int }_0^{\mathrm{\pi }/2}si{n}^{2}xdx={\int }_0^{\mathrm{\pi }/2}co{s}^{2}xdx$.

Obtain the averages of  sin2(nπx/l) and cos2(nπx/l)  

b).

From the given information, averages of$i{n}^2(n\pi x/l)andco{s}^2(n\pi x/l)$ are the same over a period is given below.

The period is as follows:

$\frac{2\pi }{T}=\frac{n\pi }{l}\Rightarrow T=\frac{2l}{n}$

Thus, $x=\frac{1}{2n}-t.$.

$dx=-dt$

Differentiate further as shown below.

$$\begin{gathered} {\int }_0^{\mathrm{\pi ll}2}si{n}^2\frac{n\pi x}{l}={\int }_{0l2n}^{-3l2n}si{n}^2\left(\frac{\pi }{2}-\frac{n\pi }{l}t\right)dt\left(-1\right) \\ ={\int }_{-3l2n}^{1l2n}{\cos }^2\frac{n\pi t}{l}dt \\ ={\int }_0^{1l2n}{\cos }^2\frac{n\pi t}{l}dt \\ ={\int }_0^{1l2n}{\cos }^2\frac{n\pi t}{l}dt \end{gathered}$$

Thus, the solution is localid="1659241525185" ${\int }_0^{\mathrm{\pi ll}2}si{n}^2\frac{n\pi x}{l}dx={\int }_0^{\mathrm{\pi ll}2}co{s}^2\frac{n\pi x}{l}dx$.