A) \[\pi \]
B) \[a\pi \]
C) \[\frac{\pi }{2}\]
D) \[2\pi \]
Correct Answer: C
Solution :
\[\int\limits_{-\pi }^{\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{x}}}}dx=I,\] Also \[I=\int\limits_{-\pi }^{\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{-x}}}}\,\,dx\] \[I+I=\int\limits_{-\pi }^{\pi }{{{\cos }^{2}}\,x\,dx=2}\int\limits_{0}^{\pi }{{{\cos }^{2}}x\,dx}\] \[\left. =2.\frac{1}{2}\int\limits_{0}^{\pi }{\left( 1+\cos 2x \right)}\,\,dx=\left( x+\frac{\sin 2x}{2} \right) \right|_{0}^{\pi }=\pi \] \[\Rightarrow \,\,I=\frac{\pi }{2}\]
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