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