Question

Evaluate: \[ \int \frac{dx}{\cos x+\sqrt{3}\sin x} \]

• A. $\log \tan\left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)+c$
• B. $\log \tan\left(\dfrac{x}{2}-\dfrac{\pi}{12}\right)+c$
• C. $\dfrac{1}{2}\log \tan\left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)+c$
• D. $\dfrac{1}{2}\log \tan\left(\dfrac{x}{2}-\dfrac{\pi}{12}\right)+c$

Hint:

Expressions of the form $a\cos x+b\sin x$ can be simplified using the identity $a\cos x+b\sin x=R\cos(x-\alpha)$, which makes integration easy.

Answer 1

The Correct Option is C

Step 1: Combine $\cos x+\sqrt{3}\sin x$ into a single trigonometric function. \[ \cos x+\sqrt{3}\sin x = 2\cos\left(x-\frac{\pi}{3}\right) \]
Step 2: Substitute in the integral: \[ \int \frac{dx}{\cos x+\sqrt{3}\sin x} = \frac{1}{2}\int \sec\left(x-\frac{\pi}{3}\right)\,dx \]
Step 3: Use the standard integral: \[ \int \sec u\,du = \log \tan\left(\frac{u}{2}+\frac{\pi}{4}\right)+c \]
Step 4: Here, $u=x-\dfrac{\pi}{3}$. \[ \frac{u}{2}+\frac{\pi}{4} = \frac{x}{2}-\frac{\pi}{6}+\frac{\pi}{4} = \frac{x}{2}+\frac{\pi}{12} \]
Step 5: Therefore, \[ \int \frac{dx}{\cos x+\sqrt{3}\sin x} = \frac{1}{2}\log \tan\left(\frac{x}{2}+\frac{\pi}{12}\right)+c \]