Question

\( \int_{0}^{\pi/2} \frac{dx}{\cos x - \sqrt{3}\sin x} = \)

• A. 0
• B. \( \frac{1}{2} \log(2-\sqrt{3}) \)
• C. \( \frac{1}{2} \log(2+\sqrt{3}) \)
• D. \( \frac{1}{2} \log(2\sqrt{3}-3) \)

Hint:

For integrals with \( a\cos x + b\sin x \) in the denominator, always convert the expression to the form \( R\cos(x \mp \alpha) \) or \( R\sin(x \pm \alpha) \). This transforms the integral into a standard form involving sec or csc.

Answer 1

The Correct Option is D

The integral is improper as the denominator becomes zero at \(x=\pi/6\). The question as stated is problematic. However, in the context of this exam, it's likely intended to be solved formally.
First, rewrite the denominator in the form \( R\cos(x+\alpha) \).
\( \cos x - \sqrt{3}\sin x \). Here \(a=1, b=-\sqrt{3}\).
\( R = \sqrt{1^2+(-\sqrt{3})^2} = \sqrt{1+3}=2 \).
\( 2(\frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x) = 2(\cos(\pi/3)\cos x - \sin(\pi/3)\sin x) = 2\cos(x+\pi/3) \).
The integral becomes \( \int_0^{\pi/2} \frac{dx}{2\cos(x+\pi/3)} = \frac{1}{2} \int_0^{\pi/2} \sec(x+\pi/3) dx \).
The integral of \( \sec u \) is \( \ln|\sec u + \tan u| \).
\( \frac{1}{2} [\ln|\sec(x+\pi/3) + \tan(x+\pi/3)|]_0^{\pi/2} \).
At the upper limit \(x=\pi/2\): \( x+\pi/3 = \pi/2+\pi/3 = 5\pi/6 \).
\( \sec(5\pi/6) = -2/\sqrt{3} \), \( \tan(5\pi/6) = -1/\sqrt{3} \). Sum is \( -3/\sqrt{3} = -\sqrt{3} \).
At the lower limit \(x=0\): \( x+\pi/3 = \pi/3 \).
\( \sec(\pi/3) = 2 \), \( \tan(\pi/3) = \sqrt{3} \). Sum is \( 2+\sqrt{3} \).
The evaluation is \( \frac{1}{2}(\ln|-\sqrt{3}| - \ln|2+\sqrt{3}|) = \frac{1}{2}(\ln(\sqrt{3}) - \ln(2+\sqrt{3})) = \frac{1}{2}\ln(\frac{\sqrt{3}}{2+\sqrt{3}}) \).
Let's simplify the argument of the log: \( \frac{\sqrt{3}(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} = \frac{2\sqrt{3}-3}{4-3} = 2\sqrt{3}-3 \).
So the result is \( \frac{1}{2}\ln(2\sqrt{3}-3) \). This matches option (D).