Question

The Integral
$\int \frac{(1 - \frac{1}{\sqrt{3}})(\cos x - \sin x)}{1 + \frac{2}{\sqrt{3}}\sin2 x} \,dx$
is equal to

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

Answer 1

Answer: (A)

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

Answer 2

The correct answer is (A) : $\frac{1}{2} \log_e \left| \frac{\tan\left(\frac{\pi}{2} + \frac{\pi}{12}\right)}{\tan\left(\frac{\pi}{2} + \frac{\pi}{6}\right)} \right| + C$
$=\int \frac{(1 - \frac{1}{\sqrt{3}})(\cos x - \sin x)}{1 + \frac{2}{\sqrt{3}}\sin2 x} \,dx$
$= ∫\frac{(\frac{\sqrt{3}-1}{\sqrt{3}}) \sqrt2\sin(\frac{π}{4}-x)}{(\frac{2}{\sqrt3})(\sin \frac{π}{3}+\sin 2x)}dx$
$= ∫ \frac{\frac{(\sqrt{3}-1)}{\sqrt2}\sin(\frac{π}{4}-x)}{\sin(\frac{π}{3}+\sin2x)}dx$
$=\int \frac{\frac{\sqrt{3}-1}{2\sqrt{2}}\sin(\frac{\pi}{4}-x)}{\sin(\frac{\pi}{6}+x)\cos(\frac{\pi}{6}-x)} \,dx$
$= \frac{1}{2} ∫ \frac{2\sin\frac{π}{12}sin(\frac{π}{4}-x) }{\sin(\frac{π}{6}+x)\cos(\frac{π}{6}-x)}dx$
$= \frac{1}{2} ∫ \frac{\cos(\frac{π}{6}-x)-\cos(\frac{π}{3}-x)}{\sin(\frac{π}{6}+x)\cos(\frac{π}{6}-x)}dx$
$=\frac{1}{2} [ ∫\cosec (\frac{π}{6}+x)dx - ∫sec (\frac{π}{6}-x)dx]$
$=\frac{1}{2}[In |\tan(\frac{π}{12}+\frac{x}{2})|- ∫\cosec (\frac{π}{3}-x)dx]$
$=\frac{1}{2} [ In|\tan(\frac{π}{12}+ \frac{π}{2})| -In| \frac{π}{6}+\frac{π}{2}|] +C$
$=\frac{1}{2} In|\frac{\tan(\frac{π}{12}+ \frac{π}{2})}{\tan(\frac{π}{6}+ \frac{π}{2})}| +C$