A) \[\pi ab\]
B) \[{{\pi }^{2}}ab\]
C) \[\frac{\pi }{ab}\]
D) \[\frac{\pi }{2ab}\]
Correct Answer: D
Solution :
\[I=\int_{0}^{\pi /2}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}.}\] Dividing the numerator and denominator by \[{{\cos }^{2}}x,\] we get \[I=\int_{0}^{\pi /2}{\frac{\frac{1}{{{\cos }^{2}}x}dx}{{{a}^{2}}+{{b}^{2}}\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}=\int_{0}^{\pi /2}{\frac{{{\sec }^{2}}x}{{{a}^{2}}+{{b}^{2}}{{\tan }^{2}}x}dx}}\]. Substituting \[b\,\,\tan x=t\] and \[b\,\,{{\sec }^{2}}x\,dx=dt\] and limit when \[x=0\], then \[t=0\] and when \[x=\frac{\pi }{2},\]then \[t=\infty ,\] therefore, \[I=\int_{0}^{\infty }{\frac{\frac{dt}{b}}{{{a}^{2}}+{{t}^{2}}}}=\frac{1}{b}\left[ \frac{1}{a}{{\tan }^{-1}}\left( \frac{t}{a} \right) \right]_{0}^{\infty }\] \[=\frac{1}{ab}\left[ {{\tan }^{-1}}\infty -{{\tan }^{-1}}0 \right]=\frac{1}{ab}\left( \frac{\pi }{2}-0 \right)=\frac{\pi }{2ab}\].
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