A) \[\frac{1}{6}{{\tan }^{-1}}(2\tan x)+C\]
B) \[{{\tan }^{-1}}(2\tan x)+C\]
C) \[\frac{1}{6}{{\tan }^{-1}}\left( \frac{2\tan x}{3} \right)+C\]
D) None of these
Correct Answer: C
Solution :
[c] \[I=\int{\frac{1}{1+3{{\sin }^{2}}x+8{{\cos }^{2}}x}dx}\] Dividing the numerator and denominator by \[{{\cos }^{2}}x,\] we get \[I=\int{\frac{{{\sec }^{2}}x}{{{\sec }^{2}}x+3{{\tan }^{2}}x+8}dx}\] \[\Rightarrow I=\int{\frac{{{\sec }^{2}}x}{1+{{\tan }^{2}}x+3{{\tan }^{2}}x+8}dx=\int{\frac{{{\sec }^{2}}x}{4{{\tan }^{2}}x+9}dx}}\]Putting \[\tan x=t\Rightarrow {{\sec }^{2}}x\,\,dx=dt,\] we get \[I=\int{\frac{dt}{4{{t}^{2}}+9}=\frac{1}{4}\int{\frac{dt}{{{t}^{2}}+{{(3/2)}^{2}}}}}\] \[=\frac{1}{4}\times \frac{1}{3/2}{{\tan }^{-1}}\left( \frac{t}{3/2} \right)+C\] \[\Rightarrow \,\,\,\,I=\frac{1}{6}{{\tan }^{-1}}\left( \frac{2t}{3} \right)+C=\frac{1}{6}{{\tan }^{-1}}\left( \frac{2\tan x}{3} \right)+C\]
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