Question

\[ \int \frac{d x}{2+\cos x}= \]

• A. \(2\mathrm{tan}^{-1}(\frac{1}{\sqrt{3}}\mathrm{tan}\frac{x}{2})+c\), where \(c\) is the constant of integration
• B. \(\frac{2}{\sqrt{3}}\mathrm{tan}^{-1}(\frac{1}{\sqrt{3}}\mathrm{tan}\frac{x}{2})+c\), where \(c\) is the constant of integration
• C. \(\frac{1}{\sqrt{3}}\mathrm{tan}^{-1}(\frac{1}{\sqrt{3}}\mathrm{tan}\frac{x}{2})+\mathrm{c}\), where c is the constant of integration
• D. \(\sqrt{3}\mathrm{tan}^{-1}(\frac{1}{\sqrt{3}}\mathrm{tan}\frac{x}{2})+c\), where \(c\) is the constant of integration

Hint:

Use the substitution t = tan(x/2) to turn cos x into a rational function of t.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For integrals of \(\dfrac{1}{a+b\cos x}\) put \(t=\tan\dfrac{x}{2}\), so \(\cos x=\dfrac{1-t^{2}}{1+t^{2}}\) and \(dx=\dfrac{2\,dt}{1+t^{2}}\).

Step 2: Substitute
\(2+\cos x=\dfrac{2(1+t^{2})+1-t^{2}}{1+t^{2}}=\dfrac{3+t^{2}}{1+t^{2}}\).

Step 3: Integrate
\(\int\dfrac{2\,dt}{3+t^{2}}=\dfrac{2}{\sqrt{3}}\tan^{-1}\dfrac{t}{\sqrt{3}}+c\).

Step 4: Back-substitute
\(=\dfrac{2}{\sqrt{3}}\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\tan\dfrac{x}{2}\right)+c\).

Final Answer:
The integral is (2/sqrt3) tan^-1((1/sqrt3) tan(x/2)) + c. \[ \boxed{\text{(B) }\dfrac{2}{\sqrt{3}}\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\tan\dfrac{x}{2}\right)+c} \]