Let $\int \frac{2-\tan x}{3+\tan x} d x=\frac{1}{2}\left(\alpha x+\log _e|\beta \sin x+\gamma \cos x|\right)+C$, where $C$ is the constant of integration. Then $\alpha+\frac{\gamma}{\beta}$ is equal to : [JEE Main 2024, 09 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: a) 4. $\int \frac{2-\tan x}{3+\tan x}dx=\int \frac{2\cos x-\sin x}{3\cos x+\sin x}dx$ $2\cos x-\sin x=A(3\cos x+\sin x)+B(\cos x-3\sin x)$ $3A+B=2$ $A–3B=–1$ $\Rightarrow \mathrm{A}=\frac{1}{2},\mathrm{B}=\frac{1}{2}$ $∴\int \frac{2\cos x-\sin x}{3\cos x+\sin x}dx$ $=\frac{1}{2}\left[\int \frac{-3 \sin x+\cos x}{3 \cos x+\sin x} d x+\int \frac{3 \cos x+\sin x}{3 \cos x+\sin x} d x\right]$ $=\frac{x}{2}+\frac{1}{2}\ln |3\cos x+\sin x|+C$ $=\frac{1}{2}(x+\ln |3\cos x+\sin x|)+C$ $=\frac{1}{2}(\alpha x+\ln |\beta \sin x+\gamma \cos x|)+C$ $\alpha =1,\beta =1,\gamma =3$ $∴\alpha +\frac{\gamma }{\beta }=1+\frac{3}{1}=4$

Let \(\int \frac{2-\tan x}{3+\tan x} d x=\frac{1}{2}\left(\alpha x+\log _e|\beta \sin x+\gamma \cos x|\right)+C\), where…

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