A) \[-\pi \]
B) 0
C) \[\pi \]
D) \[2\pi \]
Correct Answer: D
Solution :
\[I=\int_{-\pi }^{\pi }{{{(\cos ax-\sin bx)}^{2}}dx}\] \[I=\int_{-\pi }^{\pi }{({{\cos }^{2}}ax+{{\sin }^{2}}bx-2\cos \,ax\,\,\sin bx)\,\,dx}\] \[I=\int_{-\pi }^{\pi }{({{\cos }^{2}}ax+{{\sin }^{2}}bx)\,\,dx}-\int_{-\pi }^{\pi }{2\cos ax\sin bx\,\,dx}\] \[I=2\int_{0}^{\pi }{({{\cos }^{2}}ax+{{\sin }^{2}}bx)\,\,dx}-0\] \[I=2\int_{0}^{\pi }{\left( \frac{1+\cos 2ax}{2}+\frac{1-\cos 2bx}{2} \right)\,dx}\] \[I=\int_{0}^{\pi \,}{\left( 2+\cos 2ax-\cos 2bx \right)\,dx}=2\pi .\]
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