Question

In the following integrals express the sines and cosines in exponential form and then integrate to show that

${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{2}\mathbf{x}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{4}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}\mathit{\pi }$

Short answer

The exponential form of the given question is, ${\int }_0^{2x}{\sin }^{2}4xdx={\int }_0^{2x}\frac{-1}{4}\left(e^{8ix}+e^{-8ix}-2\right)dx$and it has been proved by integration.

Step-by-step solution

Given Information.

The given equation is ${\int }_0^{2x}{\sin }^{2}4xdx=\mathrm{\pi }$.

Step 2: Meaning of exponential form.

Representing the complex number in exponential form means writing the given complex number in the form of ${\mathbf{e}}^{\mathbf{i\theta }}$

Step 3: Substitute the value in the formula to convert it in exponential form.

Consider the function

${\int }_0^{2\mathrm{\pi }}{\sin }^{2}4xdx$

Substitute the exponential form of sine in above function $\sin \theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}$.

$$\begin{gathered} \sin 4x=\frac{e^{4ix}-e^{-4ix}}{2i} \\ {\sin }^{2}4x={\left(\frac{e^{4ix}-e^{-4ix}}{2i}\right)}^2 \\ {\sin }^{2}4x=-\frac{1}{4}\left(e^{8ix}-e^{-8ix}-2\right) \end{gathered}$$

Step 4: Integrate the function.

Integrate the derived exponential function.

${\int }_0^{2\mathrm{\pi }}{\sin }^{2}4xdx={\int }_0^{2\mathrm{\pi }}\frac{-1}{4}\left(e^{8ix}+e^{-8ix}-2\right)dx$

Substitute the limit.

$$\begin{gathered} {\int }_0^{2\mathrm{\pi }}{\sin }^{2}4xdx=-\frac{1}{4}{\left(\frac{e^{8ix}}{8i}+\frac{e^{-8ix}}{-8i}-2x\right)}_0^{2\mathrm{\pi }} \\ =-\frac{1}{4}\left(\frac{e^{16ix}}{8i}+\frac{e^{-16ix}}{-8i}-4x\right)+\frac{1}{4}\left(\frac{e^0}{8i}+\frac{e^0}{-8i}-2\left(0\right)\right) \end{gathered}$$

$$\begin{gathered} =\frac{-1}{4}\left(\frac{1}{8i}\left(e^{16i\mathrm{\pi }}-e^{-16i\mathrm{\pi }}\right)-4\mathrm{\pi }\right)-0 \\ =\frac{-1}{4}\left(\frac{1}{8i}\left(2i\sin \left(16\mathrm{\pi }\right)\right)\right)+\mathrm{\pi } \\ =\frac{-1}{4}\left(\frac{1}{8\mathrm{i}}\left(2\mathrm{i}\times 0\right)\right)+\mathrm{\pi } \\ =0+\mathrm{\pi } \end{gathered}$$

${\int }_0^{2\mathrm{\pi }}{\sin }^{2}4xdx=\mathrm{\pi }$

Therefore, it has been shown that ${\int }_0^{2\mathrm{\pi }}{\sin }^{2}4xdx=\mathrm{\pi }$after integrating it in exponential form .

${\int }_0^{2\mathrm{\pi }}\frac{-1}{4}\left(e^{8ix}+e^{-8ix}-2\right)dx$.