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