Question

If the integral of $(\sin 2\mathrm{x}-\cos 2\mathrm{x})=\frac{1}{\sqrt{2}}\sin (2\mathrm{x}-\mathrm{a})+\mathrm{b}$ then find $\mathrm{a}$ and $\mathrm{b}$.

Answer 1

Step 1: Find integration of $\sin 2\mathrm{x}-\cos 2\mathrm{x}$:

$$\begin{gathered} \int (\sin 2\mathrm{x}-\cos 2\mathrm{x})\mathrm{dx}=\frac{-1}{2}\cos 2\mathrm{x}-\frac{\sin 2\mathrm{x}}{2}+\mathrm{C}\mathbf{[}\mathbf{\because }\mathbf{\int }\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{x}\mathbf{+}\mathbf{C}\mathbf{}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{}\mathbf{\int }\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{x}\mathbf{=}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}\mathbf{+}\mathbf{C}\mathbf{]} \\ =-\frac{1}{2}\left(\cos 2\mathrm{x}-\sin 2\mathrm{x}\right)+\mathrm{C} \\ =-\frac{\sqrt{2}}{2\sqrt{2}}\left(\cos 2\mathrm{x}-\sin 2\mathrm{x}\right)+\mathrm{C}\mathbf{[}\mathbf{M}\mathbf{u}\mathbf{l}\mathbf{t}\mathbf{i}\mathbf{p}\mathbf{l}\mathbf{y}\mathbf{}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{}\mathbf{d}\mathbf{i}\mathbf{v}\mathbf{i}\mathbf{d}\mathbf{e}\mathbf{}\mathbf{b}\mathbf{y}\mathbf{}\sqrt{\mathbf{2}}\mathbf{]} \\ =\frac{\sqrt{2}}{2}\left(\cos 2\mathrm{x}\left(-\frac{1}{\sqrt{2}}\right)+\sin 2\mathrm{x}\left(-\frac{1}{\sqrt{2}}\right)\right)+\mathrm{C} \\ =\frac{1}{\sqrt{2}}\left(\cos 2\mathrm{x}·\sin \frac{5\mathrm{\pi }}{4}+\sin 2\mathrm{x}·\cos \frac{5\mathrm{\pi }}{4}\right)+\mathrm{C}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{a}\mathbf{r}\mathbf{e}\mathbf{}\mathbf{n}\mathbf{e}\mathbf{g}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{I}\mathbf{I}\mathbf{I}\mathbf{}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{d}\mathbf{r}\mathbf{a}\mathbf{n}\mathbf{t}\mathbf{]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=\frac{1}{\sqrt{2}}\left(\sin \left(2\mathrm{x}+\frac{5\mathrm{\pi }}{4}\right)\right)+\mathrm{C}\mathbf{\left[}\mathbf{\because }\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{=}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{a}\mathbf{·}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{b}\mathbf{+}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{a}\mathbf{·}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{b}\mathbf{\right]} \end{gathered}$$

Step 2: Compare with given equation to find $a$ and $b$

Comparing $\frac{1}{\sqrt{2}}\left(\sin \left(2\mathrm{x}+\frac{5\mathrm{\pi }}{4}\right)\right)+\mathrm{C}\mathrm{and}\frac{1}{\sqrt{2}}\sin (2\mathrm{x}-\mathrm{a})+\mathrm{b}$:

After comparing we get the values of a and b.

$\mathrm{a}=\frac{-5\mathrm{\pi }}{4}\mathrm{and}\mathrm{b}=\mathrm{C}\left(\mathrm{constant}\right),\mathrm{b}\in \mathrm{all}\mathrm{real}\mathrm{numbers}$

Therefore $\mathrm{a}=\frac{-5\mathrm{\pi }}{4}\mathrm{and}\mathrm{b}\in \mathrm{R}$ (all real numbers).