Integral Calculus Question 55
Question: $ \int_{{}}^{{}}{\frac{dx}{1+3{{\sin }^{2}}x}=} $
[Roorkee 1989; DCE 2001]
Options:
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
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Answer:
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}+{{( \frac{1}{2} )}^{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. $
BETA
