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

(A) $\frac{1}{2} \log _{e}\left|\frac{\tan \left(\frac{x}{2}+\frac{\pi}{12}\right)}{\tan\left(\frac{x}{2}+\frac{\pi}{6}\right)}\right|+C$

(B) $\frac{1}{2} \log _{e}\left|\frac{\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)}{\tan\left(\frac{x}{2}+\frac{\pi}{3}\right)}\right|+C$

(C) $\log _{e}\left|\frac{\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)}{\tan \left(\frac{x}{2}+\frac{\pi}{12}\right)}\right|+C$

(D) $\frac{1}{2} \log _{e}\left|\frac{\tan \left(\frac{x}{2}-\frac{\pi}{12}\right)}{\tan \left(\frac{x}{2}-\frac{\pi}{6}\right)}\right|+C$

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Indefinite Integrals

The integral

\int \frac{( 1 - \frac{1}{3} ) ( c o s x - s i n x )}{( 1 + \frac{2}{3} s i n 2 x )} d x

is equal to

(A) $\frac{1}{2} lo g_{e} \frac{t a n ( \frac{x}{2} + \frac{π}{12} )}{t a n ( \frac{x}{2} + \frac{π}{6} )} + C$ ✓ Correct
(B) $\frac{1}{2} lo g_{e} \frac{t a n ( \frac{x}{2} + \frac{π}{6} )}{t a n ( \frac{x}{2} + \frac{π}{3} )} + C$
(C) $lo g_{e} \frac{t a n ( \frac{x}{2} + \frac{π}{6} )}{t a n ( \frac{x}{2} + \frac{π}{12} )} + C$
(D) $\frac{1}{2} lo g_{e} \frac{t a n ( \frac{x}{2} - \frac{π}{12} )}{t a n ( \frac{x}{2} - \frac{π}{6} )} + C$

Answer: (A)

Rewrite the denominator:

1 + \frac{2}{3} sin 2 x = \frac{3}{2} + sin 2 x = 2 sin (x + \frac{π}{6}) cos (x - \frac{π}{6})

. The numerator

(1 - \frac{1}{3}) (cos x - sin x) = cos (x - \frac{π}{6}) - sin (x + \frac{π}{6})

(after simplification). The integral becomes

\frac{1}{2} \int (\frac{1}{s i n ( x + \frac{π}{6} )} - \frac{1}{c o s ( x - \frac{π}{6} )}) d x = \frac{1}{2} ln \frac{t a n ( \frac{x}{2} + \frac{π}{12} )}{t a n ( \frac{x}{2} + \frac{π}{6} )} + C

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option