A) \[\frac{1}{3}{{\tan }^{-1}}(3{{\tan }^{2}}x)+c\]
B) \[\frac{1}{2}{{\tan }^{-1}}(2\tan x)+c\]
C) \[{{\tan }^{-1}}(\tan x)+c\]
D) None of these
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{dx}{1+3{{\sin }^{2}}x}}=\int_{{}}^{{}}{\frac{dx}{{{\sin }^{2}}x+{{\cos }^{2}}x+3{{\sin }^{2}}x}}\] \[=\int_{{}}^{{}}{\frac{dx}{4{{\sin }^{2}}x+{{\cos }^{2}}x}}=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x\,dx}{4{{\tan }^{2}}x+1}=\frac{1}{4}\int_{{}}^{{}}{\frac{{{\sec }^{2}}x\,dx}{{{\tan }^{2}}x+\frac{1}{4}}}}\] Put \[t=\tan x\Rightarrow dt={{\sec }^{2}}x\,dx,\] then it reduces to \[\frac{1}{4}\int_{{}}^{{}}{\frac{dt}{{{t}^{2}}+{{\left( \frac{1}{2} \right)}^{2}}}}=\frac{1}{4}2{{\tan }^{-1}}(2t)+c\] \[=\frac{1}{2}{{\tan }^{-1}}(2t)+c=\frac{1}{2}{{\tan }^{-1}}(2\tan x)+c.\]
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