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