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

${\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{\pi }}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{3}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}\mathit{\pi }$

The given equation is ${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}co{s}^{2}3xdx=\pi$.

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

Consider the function

${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}co{s}^{2}3xdx$

Substitute the sine and cosines in exponential form in above function as .

role="math" localid="1658748384761" $$\begin{gathered} \cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2} \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}co{s}^{2}3xdx={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{\left(\cos 3x\right)}^{2}dx \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}co{s}^{2}3xdx={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{\left(\frac{e^{3ix}+e^{-3ix}}{2}\right)}^{2}dx \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}co{s}^{2}3xdx=\frac{1}{4}{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{\left(e^{3ix}+e^{-3ix}\right)}^{2}dx \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}co{s}^{2}3xdx=\frac{1}{4}{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left(e^{6ix}+e^{-6ix}+2\right)dx \end{gathered}$$

Integrate the derived exponential function.

${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}co{s}^{2}3xdx=\frac{1}{4}{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left(e^{6ix}+e^{-6ix}+2\right)dx$

Substitute the limit.

$$\begin{gathered} {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}co{s}^{2}3xdx=\frac{1}{4}{\left[\frac{e^{6ix}}{6i}+\frac{e^{-6ix}}{-6i}+2x\right]}_{-\pi }^{\pi } \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}co{s}^{2}3xdx=\frac{1}{4}\left[\left(\frac{1}{6i}\left(e^{6ix}-e^{-6ix}\right)\right)+2\pi \left(\frac{1}{6i}\left(e^{-6ix}-e^{6ix}\right)\right)-2\pi \right] \\ =\frac{1}{4}\left[\left(\frac{1}{6i}{e}^{6ix}-e^{-6ix}{e}^{-6ix}+e^{6ix}\right)\right] \\ =\left[\left(\frac{1}{6i}\left(\frac{e^{6ix}-e^{-6ix}}{2i}\right)\right)+\pi \right] \\ =\frac{1}{6}\sin 6\pi +\pi \\ =0+\pi \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}co{s}^{2}3xdx=\pi \end{gathered}$$

Therefore, it has been shown that${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}co{s}^{2}3xdx=\pi$ after integrating it in exponential form $\frac{1}{4}{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left(e^{6ix}+e^{-6ix}+2\right)dx$.