In the following integrals express the sines and cosines in exponential form and then integrate to show that ${\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{\pi }}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{x}\mathit{c}\mathit{o}\mathit{s}\mathbf{3}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}\mathbf{0}$

The given equation is ${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\cos 2\mathrm{xcos}3\mathrm{xdx}=0$.

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

${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}cos2xcos3xdx$

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

$$\begin{gathered} \cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}. \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}cos2xcos3xdx={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left(\frac{e^{i2x}+e^{-i2x}}{e}\right)\left(\frac{e^{i3x}+e^{-i3x}}{2}\right)dx \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}cos2xcos3xdx=\frac{1}{4}{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left(e^{i2x}+e^{-i2x}\right)\left(e^{i3x}+e^{-i3x}\right)dx \\ {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}cos2xcos3xdx=\frac{1}{4}{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left(e^{i5x}+e^{ix}+e^{-ix}+e^{-i5x}\right)dx \end{gathered}$$

Integrate the derived exponential function.

$$\begin{gathered} \frac{1}{4}{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left(e^{i5x}+e^{ix}+e^{-ix}+e^{-5x}\right)dx=\frac{1}{4}{\left[\frac{e^{i5x}}{5i}+\frac{e^{ix}}{i}+\frac{e^{ix}}{-1}+\frac{e^{-i5x}}{-5i}\right]}_{-\mathrm{\pi }}^{\mathrm{\pi }} \\ \frac{1}{4}{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left({\mathrm{e}}^{\mathrm{i}5\mathrm{x}}+{\mathrm{e}}^{\mathrm{ix}}+{\mathrm{e}}^{-\mathrm{ix}}+{\mathrm{e}}^{-5\mathrm{x}}\right)\mathrm{dx}=\frac{1}{4}{\left[\frac{{\mathrm{e}}^{\mathrm{i}5\mathrm{x}}}{5\mathrm{i}}+\frac{{\mathrm{e}}^{\mathrm{ix}}}{\mathrm{i}}+\frac{{\mathrm{e}}^{\mathrm{ix}}}{-1}+\frac{{\mathrm{e}}^{-\mathrm{i}5\mathrm{x}}}{-5\mathrm{i}}\right]}_{-\mathrm{\pi }}^{\mathrm{\pi }} \end{gathered}$$

Substitute the limit.

$$\begin{gathered} \frac{1}{4}{\left[\frac{{\mathrm{e}}^{\mathrm{i}5\mathrm{x}}}{5i}+\frac{{\mathrm{e}}^{\mathrm{ix}}}{i}-\frac{e^{-\mathrm{ix}}}{i}-\frac{{\mathrm{e}}^{-5\mathrm{x}}}{5i}\right]}_{-\mathrm{\pi }}^{\mathrm{\pi }}=\frac{1}{4}\left[\frac{e^{i\mathrm{\pi }}-e^{-i5\mathrm{\pi }}}{5i}+\frac{e^{i\mathrm{\pi }}-e^{-i\mathrm{\pi }}}{i}-\frac{e^{-i\mathrm{\pi }}-e^{i\mathrm{\pi }}}{i}-\frac{e^{-i5\mathrm{\pi }}-e^{i5\mathrm{\pi }}}{5i}\right] \\ =\frac{1}{4}\left[\frac{2}{5i}\left(e^{i5\mathrm{\pi }}\right)+\frac{2}{i}{e}^{i\mathrm{\pi }}-\frac{2}{i}{e}^{-\mathrm{\pi }}-\frac{2}{5i}{e}^{-i{5}^{\mathrm{\pi }}}\right] \\ =\left[\frac{1}{5}\left(\frac{e^{i5\mathrm{\pi }}-e^{-i5\mathrm{\pi }}}{2i}\right)+\left(\frac{e^{i\mathrm{\pi }}-e^{i\mathrm{\pi }}}{2i}\right)\right] \\ =\left[\frac{1}{5}\sin 5\pi +\mathrm{sin\pi }\right] \\ \frac{1}{4}{\left[\frac{{\mathrm{e}}^{\mathrm{i}5\mathrm{x}}}{5i}+\frac{{\mathrm{e}}^{\mathrm{ix}}}{i}-\frac{e^{-\mathrm{ix}}}{i}-\frac{{\mathrm{e}}^{-5\mathrm{x}}}{5i}\right]}_{-\mathrm{\pi }}^{\mathrm{\pi }}=0 \end{gathered}$$

Therefore, it has been shown that ${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\cos 2\mathrm{xcos}3\mathrm{xdx}=0$after integrating it in exponential form $\frac{1}{4}{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left({\mathrm{e}}^{\mathrm{i}5\mathrm{x}}+{\mathrm{e}}^{\mathrm{ix}}+{\mathrm{e}}^{-\mathrm{ix}}+{\mathrm{e}}^{-5\mathrm{x}}\right)dx$.