Integral Calculus Question 51
Question: $ \int_{{}}^{{}}{\frac{dx}{5+4\cos x}=} $
[Roorkee 1983; RPET 1997]
Options:
A) $ \frac{2}{3}{{\tan }^{-1}}( \frac{1}{3}\tan x )+c $
B) $ \frac{1}{3}{{\tan }^{-1}}( \frac{1}{3}\tan x )+c $
C) $ \frac{2}{3}{{\tan }^{-1}}( \frac{1}{3}\tan \frac{x}{2} )+c $
D) $ \frac{1}{3}{{\tan }^{-1}}( \frac{1}{3}\tan \frac{x}{2} )+c $
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Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{dx}{5+4\cos x}} $ $ =\int_{{}}^{{}}{\frac{dx}{5+4[ \frac{1-{{\tan }^{2}}\frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}} ]}}=\int_{{}}^{{}}{\frac{{{\sec }^{2}}\frac{x}{2}}{9+{{\tan }^{2}}\frac{x}{2}}},dx $ Put $ \tan \frac{x}{2}=t, $ then it reduces to $ 2\int_{{}}^{{}}{\frac{dt}{3^{2}+t^{2}}=\frac{2}{3}{{\tan }^{-1}}[ \frac{1}{3}\tan \frac{x}{2} ]+c} $ Aliter : Apply direct formula i.e., $ \int_{{}}^{{}}{\frac{1}{a+b\cos x},dx} $ , {a > b} $ =\frac{2}{\sqrt{a^{2}-b^{2}}}{{\tan }^{-1}}[ \sqrt{\frac{a-b}{a+b}}\tan \frac{x}{2} ]+c $ We get $ \int_{{}}^{{}}{\frac{dx}{5+4\cos x}}=\frac{2}{3}{{\tan }^{-1}}{ \frac{1}{3}\tan \frac{x}{2} }+c. $
BETA
