A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{3}\]
D) None of these
Correct Answer: A
Solution :
Divide \[{{N}^{r}}\] and \[{{D}^{r}}\]by \[{{\cos }^{4}}x\] \ \[I=\int_{0}^{\pi /4}{\frac{{{\sec }^{2}}x{{\sec }^{2}}x\,dx}{1-{{\tan }^{2}}x+{{\tan }^{4}}x}}\] Put \[\tan x=t\]and \[{{\sec }^{2}}xdx=dt\]and adjust the limits, we get \[I=\int_{0}^{1}{\frac{(1+{{t}^{2}})}{{{t}^{4}}-{{t}^{2}}+1}}\,dt\] \[=\left[ {{\tan }^{-1}}\frac{{{t}^{2}}-1}{t} \right]_{0}^{1}={{\tan }^{-1}}(0)-{{\tan }^{-1}}(-\infty )=\frac{\pi }{2}\].
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