A) \[\frac{1}{\sqrt{2}}{{\tan }^{-1}}(\sqrt{2}\tan x)+k\]
B) \[\sqrt{2}{{\tan }^{-1}}(\sqrt{2}\tan x)+k\]
C) \[-\frac{1}{\sqrt{2}}{{\tan }^{-1}}(\sqrt{2}\tan x)+k\]
D) \[-\sqrt{2}{{\tan }^{-1}}(\sqrt{2}\tan x)+k\]
Correct Answer: A
Solution :
\[I=\int_{{}}^{{}}{\frac{1}{1+{{\sin }^{2}}x}\,dx}=\int_{{}}^{{}}{\frac{dx}{2{{\sin }^{2}}x+{{\cos }^{2}}x}}\] \[=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x\,dx}{2{{\tan }^{2}}x+1}}\]\[=\frac{1}{2}\int_{{}}^{{}}{\frac{{{\sec }^{2}}x\,dx}{{{\tan }^{2}}x+\frac{1}{2}}}\] Put \[\tan x=t\Rightarrow {{\sec }^{2}}x\,dx=dt,\] then \[I=\frac{1}{2}\int_{{}}^{{}}{\frac{dt}{{{t}^{2}}+\frac{1}{2}}=\frac{1}{2}}.\frac{1}{1/\sqrt{2}}{{\tan }^{-1}}\frac{t}{1/\sqrt{2}}\] \[=\frac{1}{\sqrt{2}}{{\tan }^{-1}}(\sqrt{2}\tan x)+k\].
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