JEE Main 2024 (Online) 9th April Morning ShiftQuestion 1
Let $\int\frac{2-\tan x}{3+\tan x}dx=\frac{1}{2}\left(\alpha x+\log_e|\beta\sin x+\gamma\cos x|\right)+C$, where $C$ is the constant of integration. Then $\alpha+\frac{\gamma}{\beta}$ is equal to :
Indefinite Integration
Solution
Concept. Write the integrand in sines and cosines and split the numerator as $A\cdot$(denominator) $+\,B\cdot$(its derivative). Formulas used. (1) $\int\dfrac{A\,D+B\,D'}{D}\,dx=Ax+B\ln|D|$Step 1 - Rewrite. Multiply top and bottom by $\cos x$: $$\frac{2-\tan x}{3+\tan x}=\frac{2\cos x-\sin x}{3\cos x+\sin x}.$$ Step 2 - Split. With $D=3\cos x+\sin x$, $D'=\cos x-3\sin x$: $$2\cos x-\sin x=A(3\cos x+\sin x)+B(\cos x-3\sin x):\quad3A+B=2,\ A-3B=-1.$$ So $A=B=\frac12$. Step 3 - Integrate. By (1), $$\int=\frac12\Big(x+\ln|\sin x+3\cos x|\Big)+C.$$ So $\alpha=1$, $\beta=1$, $\gamma=3$. Step 4 - Value. $\alpha+\frac\gamma\beta=4$. Answer: (C)