Q16P - Mathematical Methods in Physical Sciences

Question

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

${\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{\pi }}\mathit{s}\mathit{i}\mathit{n}\mathbf{3}\mathit{x}\mathbf{ }\mathit{c}\mathit{o}\mathit{s}\mathbf{ }\mathbf{4}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The exponential form of the given question is, ${\int }_{-\pi }^{\pi }\sin 3x \cos 4xdx={\int }_{-\pi }^{\pi }\left(\frac{e^{i7x}+e^{-ix}-e^{ix}-e^{-i7x}}{}\right)dx$ and it has been proved by integration.

Step 1: Given Information.

The given equation is ${\int }_{-\pi }^{\pi }\sin 3x \cos 4xdx=0$.

Step 2: Meaning of exponential form.

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

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

Consider the function

${\int }_{-\pi }^{\pi }\sin 3x \cos 4xdx$

 

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

$$\begin{gathered} \left(\sin \theta ,\cos \theta \right)=\left(\frac{e^{i\theta }-e^{-i\theta }}{2i},\frac{e^{i\theta }+e^{-i\theta }}{2}\right) \\ \sin 3x \cos 4x=\left(\frac{e^{i3x}-e^{-i3x}}{2i}\right)\left(\frac{e^{i4x}+e^{-i4x}}{2}\right) \\ \sin 3x \cos 4x=\frac{e^{i3x}.e^{i4x}+e^{i3x}.e^{-i4x}-e^{-i3x}.e^{i4x}-e^{-i3x}.e^{-i4x}}{4i} \\ \sin 3x \cos 4x=\frac{e^{i7x}+e^{-ix}-e^{ix}-e^{-i7x}}{4i} \end{gathered}$$

Step 4: Integrate the function.

Integrate the derived exponential function.

${\int }_{-\pi }^{\pi }\sin 3x \cos 4xdx=\frac{1}{4i}{\int }_{-x}^{\pi }\left(e^{i7x}+e^{-ix}-e^{ix}-e^{-i7x}\right)dx$

 

Substitute the limit.

$$\begin{gathered} {\int }_{-\pi }^{\pi }\sin 3x \cos 4xdx=\frac{1}{4i}{\int }_{-\pi }^{\pi }\left(e^{i7x}+e^{-ix}-e^{ix}-e-i7x\right)dx \\ {\int }_{-\pi }^{\pi }\sin 3x \cos 4xdx=\frac{1}{4i}\left\{{\int }_{-\pi }^{\pi }{e}^{i7x}dx+{\int }_{-\pi }^{\pi }{e}^{-ix}dx-{\int }_{-\pi }^{\pi }{e}^{-i7x}dx\right\} \\ {\int }_{-\pi }^{\pi }\sin 3x \cos 4xdx=\frac{1}{4i}\left\{{\left[\frac{e^{i7x}}{7i}\right]}_{-\pi }^{\pi }+{\left[-\frac{e^{-ix}}{i}\right]}_{-\pi }^{\pi }-{\left[\frac{e^{ix}}{i}\right]}_{-\pi }^{\pi }-{\left[-\frac{e^{-i7x}}{7i}\right]}_{-\pi }^{\pi }\right\} \\ {\int }_{-\pi }^{\pi }\sin 3x \cos 4xdx=\frac{1}{4i}\left\{\frac{e^{i7\pi }}{7i}-\frac{e^{-i7\pi }}{7i}-\frac{e^{-i\pi }}{i}+\frac{e^{-i\pi }}{i}-\frac{e^{i\pi }}{i}+\frac{e^{-i\pi }}{i}+\frac{e^{i7\pi }}{7i}-\frac{e^{i7\pi }}{7i}\right\} \\ {\int }_{-\pi }^{\pi }\sin 3x \cos 4xdx=\frac{1}{4i}\left(0\right) \\ {\int }_{-\pi }^{\pi }\sin 3x \cos 4xdx=0 \end{gathered}$$

 

Therefore, it has been shown that ${\int }_{-\pi }^{\pi }\sin 3x \cos 4xdx=0$ after integrating it in exponential form ${\int }_{-\pi }^{\pi }\left(\frac{e^{i7x}+e^{-ix}-e^{ix}-e^{-i7x}}{4i}\right)dx$.