Integral Calculus Question 33
Question: $ \int_{{}}^{{}}{\frac{dx}{2+\cos x}=} $
Options:
A) $ 2{{\tan }^{-1}}( \frac{1}{\sqrt{3}}\tan \frac{x}{2} )+c $
B) $ \frac{2}{\sqrt{3}}{{\tan }^{-1}}( \frac{1}{\sqrt{3}}\tan \frac{x}{2} )+c $
C) $ \frac{1}{\sqrt{3}}{{\tan }^{-1}}( \frac{1}{\sqrt{3}}\tan \frac{x}{2} )+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{dx}{2+\cos x}=\int_{{}}^{{}}{\frac{dx}{2{{\sin }^{2}}( \frac{x}{2} )+2{{\cos }^{2}}( \frac{x}{2} )+{{\cos }^{2}}( \frac{x}{2} )-{{\sin }^{2}}( \frac{x}{2} )}}} $
$ =\int_{{}}^{{}}{\frac{dx}{{{\sin }^{2}}( \frac{x}{2} )+3{{\cos }^{2}}( \frac{x}{2} )}}=\int_{{}}^{{}}{\frac{{{\sec }^{2}}( \frac{x}{2} )}{{{\tan }^{2}}( \frac{x}{2} )+3}dx} $
Put $ \tan ( \frac{x}{2} )=t\Rightarrow {{\sec }^{2}}( \frac{x}{2} ),dx=2dt, $ then it reduces to
$ 2\int_{{}}^{{}}{\frac{dt}{t^{2}+3}}=\frac{2}{\sqrt{3}}{{\tan }^{-1}}( \frac{t}{\sqrt{3}} )+c=\frac{2}{\sqrt{3}}{{\tan }^{-1}}( \frac{\tan ( \frac{x}{2} )}{\sqrt{3}} )+c $ .
BETA
