Integral Calculus Question 167
Question: Evaluate: $ \int{\frac{1}{1+3{{\sin }^{2}}x+8{{\cos }^{2}}x}dx} $
Options:
A) $ \frac{1}{6}{{\tan }^{-1}}(2\tan x)+C $
B) $ {{\tan }^{-1}}(2\tan x)+C $
C) $ \frac{1}{6}{{\tan }^{-1}}( \frac{2\tan x}{3} )+C $
D) None of these
Show Answer
Answer:
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}}xdx=dt, $ we get $ I=\int{\frac{dt}{4t^{2}+9}=\frac{1}{4}\int{\frac{dt}{t^{2}+{{(3/2)}^{2}}}}} $ $ =\frac{1}{4}\times \frac{1}{3/2}{{\tan }^{-1}}( \frac{t}{3/2} )+C $
$ \Rightarrow I=\frac{1}{6}{{\tan }^{-1}}( \frac{2t}{3} )+C=\frac{1}{6}{{\tan }^{-1}}( \frac{2\tan x}{3} )+C $
BETA
