Question

The value of $ \int_{-\pi }^{\pi }{{{(\cos px-\sin qx)}^{2}}}dx, $ where p and q are integers, is

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

Answer 1

Answer: (D)

Let $ I=\int_{-\pi }^{\pi }{{{(\cos px-\sin qx)}^{2}}dx} $
$ =\int_{-\pi }^{\pi }{({{\cos }^{2}}px+{{\sin }^{2}}qx-2\cos px\sin qx)dx} $
$ =2\int_{0}^{\pi }{({{\cos }^{2}}px+{{\sin }^{2}}qx)dx} $
$ -\int_{-\pi }^{\pi }{2\cos px\sin qxdx} $
$ =2\int_{0}^{\pi }{({{\cos }^{2}}px+{{\sin }^{2}}qx)dx-0} $
$ \left[ \because \int_{-\pi }^{\pi }{2\cos px\sin qx\,dx\,is\,an\,odd\,function} \right] $
$ =2\int_{0}^{\pi }{\left[ \frac{(1+\cos 2px)}{2}+\frac{1}{2}(1-\cos 2xqx) \right]}dx $ $ =\int_{0}^{\pi }{[1+\cos 2px+1-\cos 2qx]}\,dx $
$ =\int_{0}^{\pi }{[2+\cos 2px-\cos 2qx]}\,dx $
$ =\left[ 2x+\frac{\sin 2px}{2p}-\frac{\sin 2qx}{2p} \right]_{0}^{\pi }=2\pi +0-0=2\pi $