Question

${\int }_{-\pi }^{\pi }(\cos px-\sin qx{)}^{2}dx$, where $p$ and $q$ are integers is equal to

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

Answer 1

The correct option is C $2\pi$

${\int }_{-\pi }^{\pi }(\cos px-\sin qx{)}^{2}dx$

$={\int }_{-\pi }^{\pi }({\cos }^{2}px+{\sin }^{2}qx-2\sin qx\cos px)dx$

$={\int }_{-\pi }^{\pi }({\cos }^{2}px+{\sin }^{2}qx)dx-2-2{\int }_{-\pi }^{\pi }\sin \left(qx\right)\cos \left(px\right)dx$

$=2{\int }_0^{\pi }({\cos }^{2}px+{\sin }^{2}qx)dx-0$

$\left\{\begin{array}{ccc}\because {\int }_{-a}^{a}f\left(x\right)dx& =2{\int }_0^{a}f\left(x\right)dx& \text{if}f(-x)=f\left(x\right)\\ & =0& \text{if}f(-x)=-f\left(x\right)\end{array}\right\}$

$=4(\frac{1}{2}.\frac{\pi }{2}+\frac{1}{2}.\frac{\pi }{2})$

$=2\pi$