Question

${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{\left[\cos px-\sin qx\right]}^{2}dx$ where $p$ and $q$ are integers, then the value of integral is

• A. $-\mathrm{\pi }$
• B. $0$
• C. $\mathrm{\pi }$
• D. $2\mathrm{\pi }$

Answer 1

The correct option is D

$2\mathrm{\pi }$

Explanation for the correct option:

Finding the value for the given integral:

Let $I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{\left[\cos px-\sin qx\right]}^{2}dx........\left(1\right)$

$$\begin{gathered} I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{\left[\cos p\left(-x\right)-\sin q\left(-x\right)\right]}^{2}dx\left(\sin ce{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx\right) \\ I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{\left[\cos p\left(x\right)+\sin q\left(x\right)\right]}^{2}dx..........\left(2\right) \end{gathered}$$

On adding $\left(1\right)$ and $\left(2\right)$, we get

$$\begin{gathered} \Rightarrow 2I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{\left[\cos p\left(x\right)-\sin q\left(x\right)\right]}^{2}dx+{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{\left[\cos p\left(x\right)+\sin q\left(x\right)\right]}^{2}dx \\ ={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}2\left[\cos p^2\left(x\right)+\sin q^2\left(x\right)\right]dx \\ =2{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left[\frac{\left(\cos 2px+1\right)}{2}+\frac{\left(1-\cos 2qx\right)}{2}\right]dx\left(\sin ce\cos 2px=2{\cos }^{2}x-1,\cos 2qx=1-2{\sin }^{2}x\right) \\ =2{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left[1+\frac{\cos 2px}{2}-\frac{\cos 2qx}{2}\right]dx \\ =2{\left[x+\frac{\sin 2px}{4}-\frac{\sin 2qx}{4}\right]}_{-\mathrm{\pi }}^{\mathrm{\pi }} \\ =2\times \left[\mathrm{\pi }+\frac{\sin 2p\mathrm{\pi }}{4}-\frac{\sin 2q\mathrm{\pi }}{4}\right]-\left[\left(-\mathrm{\pi }\right)+\frac{\sin 2p\left(-\mathrm{\pi }\right)}{4}-\frac{\sin 2q\left(-\mathrm{\pi }\right)}{4}\right] \\ \Rightarrow 2I=2\times 2\mathrm{\pi } \\ \Rightarrow I=2\mathrm{\pi } \end{gathered}$$

Thus, option (D) is the correct answer.