91Ó°ÊÓ

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

Representing the complex number in exponential form means writing the given complex number in the form of ${\mathbf{e}}^{\mathbf{¾\pm θ}}$

Consider the function

${∫}_0^{2\mathrm{π}}{\sin }^{2}4xdx$

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

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

Integrate the derived exponential function.

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

Substitute the limit.

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

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

${∫}_0^{2\mathrm{π}}{\sin }^{2}4xdx=\mathrm{π}$

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

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