Q13P - Mathematical Methods in Physical Sciences

Question

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

${\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{\pi }}\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{x}\mathit{s}\mathit{i}\mathit{n}\mathbf{3}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The exponential form of the given question ${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}sin2xsin3xdx={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}-\frac{1}{4}\left(e^{5ix}-e^{-ix}-e^{ix}+e^{-5ix}\right)dx$ and it has been proved by integration.

Step 1: Given Information.

The given equation is ${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}sin2xsin3xdx=0$.

Step 2: Meaning of exponential form.

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

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

Consider the function

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

 

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

$$\begin{gathered} \sin 2x\sin 3x=\left(\frac{e^{2ix}-e^{-2ix}}{2i}\right)\left(\frac{e^{3ix}-e^{-3ix}}{2i}\right) \\ \sin 2x\sin 3x=\frac{1}{-4}\left(e^{5ix}-e^{-ix}-e^{ix}+e^{-5ix}\right) \end{gathered}$$ 

Step 4: Integrate the function.

Integrate the derived exponential function.

${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin 2x\sin 3xdx={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{1}{-4}\left(e^{5ix}-e^{-ix}-e^{ix}+e^{-5ix}\right)dx$

 

Substitute the limit.

$$\begin{gathered} {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin 2x\sin 3xdx=\frac{1}{-4}{\left(\frac{e^{5ix}}{5i}-\frac{e^{ix}}{-i}-\frac{e^{ix}}{i}+\frac{e^{-5ix}}{-5i}\right)}_{x=-\mathrm{\pi }}^{\mathrm{\pi }} \\ =\frac{1}{-4}\left(\frac{e^{i5\mathrm{\pi }}-e^{-5i\mathrm{\pi }}}{5i}+\frac{e^{-i\mathrm{\pi }}-e^{-i\mathrm{\pi }}}{i}\right)-\frac{1}{-4}\left(\frac{e^{-i5\mathrm{\pi }}-e^{-5i\mathrm{\pi }}}{5i}+\frac{e^{i\mathrm{\pi }}-e^{-i\mathrm{\pi }}}{i}\right) \\ =\frac{1}{-4}\left(\frac{2}{5i}\left(e^{i5\mathrm{\pi }}-e^{-5i\mathrm{\pi }}\right)-\frac{2}{i}\left(e^{i\mathrm{\pi }}-e^{-i\mathrm{\pi }}\right)\right) \\ =-\frac{1}{5}\left(\frac{e^{i5\mathrm{\pi }}-e^{-i\mathrm{\pi }}}{2i}\right)+\frac{\left(e^{-i\mathrm{\pi }}-e^{-i\mathrm{\pi }}\right)}{2i} \\ =-\frac{1}{5}\sin 5\mathrm{\pi }+\mathrm{sin\pi } \end{gathered}$$

${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin 2x\sin 3xdx=0$

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