A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{2a}\]
C) \[\frac{\pi }{a}\]
D) \[\frac{1}{2a}\]
Correct Answer: B
Solution :
\[=\int_{0}^{\infty }{\frac{dx}{({{a}^{2}}+{{x}^{2}})}}=\left[ \frac{1}{a}{{\tan }^{-1}}\frac{x}{a} \right]_{0}^{\infty }\] \[=\frac{1}{a}\{{{\tan }^{-1}}(\infty )-{{\tan }^{-1}}(0)\}\] \[=\frac{1}{a}\left( \frac{\pi }{2}-0 \right)=\frac{\pi }{2a}\]
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