A) \[\frac{1}{\sqrt{2}}{{\tan }^{-1}}(\tan x)+c\]
B) \[\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{1}{2}\tan x \right)+c\]
C) \[\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{1}{\sqrt{2}}\tan x \right)+c\]
D) None of these
Correct Answer: C
Solution :
\[\int_{{}}^{{}}{\frac{dx}{1+{{\cos }^{2}}x}}=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x\,dx}{{{\sec }^{2}}x+1}}=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x}{{{\tan }^{2}}x+2}}\,dx\] \[=\int_{{}}^{{}}{\frac{dt}{{{t}^{2}}+2}=\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{t}{\sqrt{2}} \right)+c}\] {Putting \[\tan x=t\}\] \[=\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{1}{\sqrt{2}}\tan x \right)+c\]. Trick : By inspection, \[\frac{d}{dx}\left\{ \frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{1}{\sqrt{2}}\tan x \right) \right\}=\frac{1}{\sqrt{2}}\left( \frac{1}{1+\frac{{{\tan }^{2}}x}{2}} \right)\frac{1}{\sqrt{2}}{{\sec }^{2}}x\] \[=\frac{1}{2}\,.\,\frac{2{{\sec }^{2}}x}{(2+{{\tan }^{2}}x)}=\frac{{{\sec }^{2}}x}{1+{{\sec }^{2}}x}=\frac{1}{1+{{\cos }^{2}}x}\].
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