A) \[\frac{2}{3}{{\tan }^{-1}}\left( \frac{1}{3}\tan x \right)+c\]
B) \[\frac{1}{3}{{\tan }^{-1}}\left( \frac{1}{3}\tan x \right)+c\]
C) \[\frac{2}{3}{{\tan }^{-1}}\left( \frac{1}{3}\tan \frac{x}{2} \right)+c\]
D) \[\frac{1}{3}{{\tan }^{-1}}\left( \frac{1}{3}\tan \frac{x}{2} \right)+c\]
Correct Answer: C
Solution :
\[\int_{{}}^{{}}{\frac{dx}{5+4\cos x}}\] \[=\int_{{}}^{{}}{\frac{dx}{5+4\left[ \frac{1-{{\tan }^{2}}\frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}} \right]}}=\int_{{}}^{{}}{\frac{{{\sec }^{2}}\frac{x}{2}}{9+{{\tan }^{2}}\frac{x}{2}}}\,dx\] Put \[\tan \frac{x}{2}=t,\] then it reduces to \[2\int_{{}}^{{}}{\frac{dt}{{{3}^{2}}+{{t}^{2}}}=\frac{2}{3}{{\tan }^{-1}}\left[ \frac{1}{3}\tan \frac{x}{2} \right]+c}\] Aliter : Apply direct formula i.e., \[\int_{{}}^{{}}{\frac{1}{a+b\cos x}\,dx}\], {a > b} \[=\frac{2}{\sqrt{{{a}^{2}}-{{b}^{2}}}}{{\tan }^{-1}}\left[ \sqrt{\frac{a-b}{a+b}}\tan \frac{x}{2} \right]+c\] We get \[\int_{{}}^{{}}{\frac{dx}{5+4\cos x}}=\frac{2}{3}{{\tan }^{-1}}\left\{ \frac{1}{3}\tan \frac{x}{2} \right\}+c.\]
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