Question

$$\int \frac{1}{\cos x + \sqrt{3}\sin x} \, dx =$$

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

Hint:

When reducing expressions like $1\cos x + \sqrt{3}\sin x$, remember that factoring out the amplitude ($2$) places it in the denominator of the full integral. This means the final solution must have a leading coefficient of $\frac{1}{2}$. This rule alone allows you to instantly eliminate choices (A) and (D)!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to calculate the indefinite integral of a rational trigonometric function fraction: $\int \frac{1}{\cos x + \sqrt{3}\sin x} \, dx$.

Step 2: Key Formula or Approach:
We simplify linear combinations of sine and cosine of the form $A\cos x + B\sin x$ by dividing and multiplying the expression by $\sqrt{A^2 + B^2}$. Here, $A = 1$ and $B = \sqrt{3}$, so: $$\sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$$ This allows us to merge the denominator terms into a single sine compound angle identity: $\sin(\theta + \phi) = \sin\theta\cos\phi + \cos\theta\sin\phi$.

Step 3: Detailed Explanation:
Let's multiply and divide the denominator expression by 2: $$\cos x + \sqrt{3}\sin x = 2 \left( \frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x \right)$$ Using standard exact trigonometric values, substitute $\frac{1}{2} = \sin\left(\frac{\pi}{6}\right)$ and $\frac{\sqrt{3}}{2} = \cos\left(\frac{\pi}{6}\right)$: $$\cos x + \sqrt{3}\sin x = 2 \left( \sin\left(\frac{\pi}{6}\right)\cos x + \cos\left(\frac{\pi}{6}\right)\sin x \right)$$ Applying the sine sum identity converts this to: $$2\sin\left(x + \frac{\pi}{6}\right)$$ Substitute this simplified expression back into our original integral: $$I = \int \frac{1}{2\sin\left(x + \frac{\pi}{6}\right)} \, dx = \frac{1}{2}\int \csc\left(x + \frac{\pi}{6}\right) \, dx$$ Recall the standard logarithmic trigonometric integration formula: $\int \csc\theta \, d\theta = \ln\left| \tan\left(\frac{\theta}{2}\right) \right| + c$. Applying this formula here gives: $$I = \frac{1}{2}\ln\left| \tan\left( \frac{x + \frac{\pi}{6}}{2} \right) \right| + c$$ Divide each term in the numerator by 2 to get the final form: $$I = \frac{1}{2}\ln\left| \tan\left( \frac{x}{2} + \frac{\pi}{12} \right) \right| + c$$ This matches option (C).

Step 4: Final Answer:
The integrated value is $\frac{1}{2}\ln\left| \tan\left(\frac{x}{2} + \frac{\pi}{12}\right) \right| + c$, which corresponds to option (C).