Question

The integral $\int \frac{(1-\frac{1}{\sqrt{3}})(\cos x - \sin x)}{1+\frac{2}{\sqrt{3}}\sin 2x} dx$ is equal to:

• A. $\frac{1}{2}\log_e \left| \frac{\tan(\frac{x}{2}+\frac{\pi}{12})}{\tan(\frac{x}{2}+\frac{\pi}{6})} \right| +C$
• B. $\frac{1}{2}\log_e \left| \frac{\tan(\frac{x}{2}+\frac{\pi}{6})}{\tan(\frac{x}{2}+\frac{\pi}{3})} \right| +C$
• C. $\log_e \left| \frac{\tan(\frac{x}{2}+\frac{\pi}{6})}{\tan(\frac{x}{2}+\frac{\pi}{12})} \right| +C$
• D. $\frac{1}{2}\log_e \left| \frac{\tan(\frac{x}{2}-\frac{\pi}{12})}{\tan(\frac{x}{2}-\frac{\pi}{6})} \right| +C$

Answer 1

Answer: (A)

$\frac{1}{2}\log_e \left| \frac{\tan(\frac{x}{2}+\frac{\pi}{12})}{\tan(\frac{x}{2}+\frac{\pi}{6})} \right| +C$

Answer 2

The given integral is $I = \int \frac{(1-\frac{1}{\sqrt{3}})(\cos x - \sin x)}{1+\frac{2}{\sqrt{3}}\sin 2x} dx$.

First, simplify the constant term: $1-\frac{1}{\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}}$. So, $I = \frac{\sqrt{3}-1}{\sqrt{3}} \int \frac{\cos x - \sin x}{1+\frac{2}{\sqrt{3}}\sin 2x} dx$.

Let's use the substitution $t = \sin x + \cos x$. Then, $dt = (\cos x - \sin x) dx$. Also, $t^2 = (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + \sin 2x$. From this, $\sin 2x = t^2 - 1$.

Substitute these into the integral: $I = \frac{\sqrt{3}-1}{\sqrt{3}} \int \frac{dt}{1+\frac{2}{\sqrt{3}}(t^2-1)}$ $I = \frac{\sqrt{3}-1}{\sqrt{3}} \int \frac{dt}{\frac{\sqrt{3}+2(t^2-1)}{\sqrt{3}}}$ $I = (\sqrt{3}-1) \int \frac{dt}{\sqrt{3}+2t^2-2}$ $I = (\sqrt{3}-1) \int \frac{dt}{2t^2 + (\sqrt{3}-2)}$ Factor out 2 from the denominator: $I = \frac{\sqrt{3}-1}{2} \int \frac{dt}{t^2 + \frac{\sqrt{3}-2}{2}}$ We can rewrite $\frac{\sqrt{3}-2}{2}$ as $-\frac{2-\sqrt{3}}{2}$. $I = \frac{\sqrt{3}-1}{2} \int \frac{dt}{t^2 - \frac{2-\sqrt{3}}{2}}$ Notice that $\frac{2-\sqrt{3}}{2} = \frac{4-2\sqrt{3}}{4} = \frac{(\sqrt{3}-1)^2}{4}$. So, let $a^2 = \frac{(\sqrt{3}-1)^2}{4}$, which means $a = \frac{\sqrt{3}-1}{2}$. The integral is of the form $\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C$. $I = \frac{\sqrt{3}-1}{2} \cdot \frac{1}{2a} \log \left| \frac{t-a}{t+a} \right| + C$ $I = \frac{\sqrt{3}-1}{2} \cdot \frac{1}{2 \cdot \frac{\sqrt{3}-1}{2}} \log \left| \frac{t - \frac{\sqrt{3}-1}{2}}{t + \frac{\sqrt{3}-1}{2}} \right| + C$ $I = \frac{\sqrt{3}-1}{2} \cdot \frac{1}{\sqrt{3}-1} \log \left| \frac{2t - (\sqrt{3}-1)}{2t + (\sqrt{3}-1)} \right| + C$ $I = \frac{1}{2} \log \left| \frac{2t - (\sqrt{3}-1)}{2t + (\sqrt{3}-1)} \right| + C$.

Now, substitute back $t = \sin x + \cos x$. $I = \frac{1}{2} \log \left| \frac{2(\sin x + \cos x) - (\sqrt{3}-1)}{2(\sin x + \cos x) + (\sqrt{3}-1)} \right| + C$.

To match the options, we need to convert this expression into terms of $\tan(\frac{x}{2}+\alpha)$. We know that $\sin x + \cos x = \sqrt{2} \sin(x+\frac{\pi}{4})$.

After applying sum-to-product formulas and simplifying, we get:

$I = \frac{1}{2} \log \left| \frac{\tan(\frac{x}{2}+\frac{\pi}{12})}{\tan(\frac{x}{2}+\frac{\pi}{6})} \right| + C$.

This matches option (A).