Question

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

${\mathbf{\int }}_{\mathbf{-}\mathbf{x}}^{\mathbf{x}}\mathbf{sin}\mathbf{2}\mathbf{xcos}\mathbf{3}\mathbf{xdx}\mathbf{=}\mathbf{0}$

Short answer

The exponential form of the given question is,

${\int }_{-x}^x\sin 2x\cos 3xdx={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left(\frac{e^{15x}+e^{-ix}-e^{ix}-e^{-15x}}{4i}\right)dx$and it has been proved by integration.

Step-by-step solution

Given Information.

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

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

${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin 2x\cos 3xdx$

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

$$\begin{gathered} \left(\sin \theta ,\cos \theta \right)=\left(\frac{e^{i\theta }-e^{-i\theta }}{2i},\frac{e^{i\theta }-e^{-i\theta }}{2i}\right) \\ \sin 2x\cos 3x=\left(\frac{e^{12x}-e^{-12x}}{2i}\right)\left(\frac{e^{13x}+e^{-13x}}{2i}\right) \\ \sin 2x\cos 3x=\frac{e^{12x}.e^{13x}+e^{12x}.e^{-13x}-e^{12x}.e^{13x}-e^{12x}.e^{-13x}}{4i} \\ \sin 2x\cos 3x=\frac{e^{15x}+e^{-ix}-e^{ix}-e^{-15x}}{4i} \end{gathered}$$

Step 4: Integrate the function.

Integrate the derived exponential function.

${\int }_{-x}^x\sin 2x\cos 3xdx=\frac{1}{4i}{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left(e^{15x}+e^{-ix}-e^{ix}-e^{-15x}\right)dx$

Substitute the limit.

${\int }_{-x}^x\sin 2x\cos 3xdx=\frac{1}{4i}\left\{{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{e}^{15x}dx+{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{e}^{ix}dx-{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{e}^{-15x}dx+\right\}$

${\int }_{-x}^x\sin 2x\cos 3xdx=\frac{1}{4i}\left\{{\left[\frac{e^{15x}}{5i}\right]}_{-x}^x+{\left[-\frac{e^{1x}}{i}\right]}_{-x}^x-{\left[\frac{e^x}{5i}\right]}_{-x}^x-{\left[\frac{e^{-15x}}{5i}\right]}_{-x}^x\right\}$

${\int }_{-x}^x\sin 2x\cos 3xdx=\frac{1}{4i}\left\{\frac{e^{15x}}{5i}-\frac{e^{-15x}}{5i}-\frac{e^{-ix}}{i}-\frac{e^{ix}}{i}-\frac{e^{-ix}}{i}+\frac{e{-}^{-15ix}}{5i}-\frac{e^{15ix}}{i}\right\}$

$$\begin{gathered} {\int }_{-x}^x\sin 2x\cos 3xdx=\frac{1}{4i}\left(0\right) \\ {\int }_{-x}^x\sin 2x\cos 3xdx=0 \end{gathered}$$

Therefore, it has been shown that ${\int }_{-x}^x\sin 2x\cos 3xdx=0$after integrating it in exponential form ${\int }_{-x}^x\left(\frac{e^{15x}+e^{-ix}-e^{ix}-e^{-15x}}{4i}\right)dx.$.