A) \[-\pi \]
B) 0
C) \[\pi \]
D) \[2\pi \]
Correct Answer: D
Solution :
\[I=\int_{-\pi }^{\pi }{({{\cos }^{2}}px+{{\sin }^{2}}qx-2\sin qx\cos px)dx}\] \[=\int_{-\pi }^{\pi }{({{\cos }^{2}}px+{{\sin }^{2}}qx)dx-2\int_{-\pi }^{\pi }{\sin qx\cos px\,dx}}\] \[=2\int_{0}^{\pi }{({{\cos }^{2}}px+{{\sin }^{2}}qx)dx-0}\] \[y=4x-{{x}^{2}}\].
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