Question

$\int \frac{dx}{2+\cos x} =$}

• A. $2\tan^{-1}\left(\frac{1}{\sqrt{3}}\tan\frac{x}{2}\right) + c$
• B. $\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{1}{\sqrt{3}}\tan\frac{x}{2}\right) + c$
• C. $\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{1}{\sqrt{3}}\tan\frac{x}{2}\right) + c$
• D. $\sqrt{3}\tan^{-1}\left(\frac{1}{\sqrt{3}}\tan\frac{x}{2}\right) + c$

Hint:

For $\int \frac{dx}{a+b\cos x}$, if $a > b$, the result involves $\tan^{-1}$.

Answer 1

The Correct Option is B

Step 1: Concept
Use the substitution $t = \tan(x/2)$. Then $\cos x = \frac{1-t^2}{1+t^2}$ and $dx = \frac{2dt}{1+t^2}$.

Step 2: Meaning
This standard substitution converts a trigonometric integral into a rational algebraic integral.

Step 3: Analysis
$\int \frac{\frac{2dt}{1+t^2}}{2 + \frac{1-t^2}{1+t^2}} = \int \frac{2dt}{2(1+t^2) + 1 - t^2} = \int \frac{2dt}{t^2 + 3}$. $= 2 \int \frac{dt}{t^2 + (\sqrt{3})^2} = 2 \cdot \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) + c$.

Step 4: Conclusion
Replacing $t$ gives $\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{1}{\sqrt{3}}\tan\frac{x}{2}\right) + c$. Final Answer: (B)