Let \int \frac2-\tan x3+\tan x ~d x=\frac12(\alpha x+\log _e|\beta \sin x+\gamma \cos x|)+C , where C is the constant of integration. Then \alpha+\frac\gamma\beta is equal to :

Indefinite Integrals PYQ — JEE Main 2024 · 9 Apr · Shift 1 · Q41 with Solution

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Correct answer: D

We need to evaluate $I=\int \frac{2-\tan x}{3+\tan x}\,dx$ and compare it with $\frac12\left(\alpha x+\log_e|\beta\sin x+\gamma\cos x|\right)+C.$
Rewrite the integrand in terms of $\sin x$ and $\cos x$ :

=\frac{2\cos x-\sin x}{3\cos x+\sin x}.$$ So, $$I=\int \frac{2\cos x-\sin x}{3\cos x+\sin x}\,dx.$$ 3. Express the numerator in terms of the derivative of the denominator. Let $$D=3\cos x+\sin x.$$ Then $$D'=-3\sin x+\cos x=\cos x-3\sin x.$$ We want constants $A,B$ such that $$2\cos x-\sin x=A(\cos x-3\sin x)+B(3\cos x+\sin x).$$ Comparing coefficients: $$A+3B=2,\qquad -3A+B=-1.$$ Solving, $$A=\frac{5}{10}=\frac12,\qquad B=\frac12.$$ Thus, $$2\cos x-\sin x=\frac12(\cos x-3\sin x)+\frac12(3\cos x+\sin x).$$ Hence, $$\frac{2\cos x-\sin x}{3\cos x+\sin x}=\frac12\frac{\cos x-3\sin x}{3\cos x+\sin x}+\frac12.$$ 4. Integrate: $$I=\frac12\int \frac{\cos x-3\sin x}{3\cos x+\sin x}\,dx+\frac12\int dx.$$ Since $\cos x-3\sin x=(3\cos x+\sin x)'$, $$\int \frac{\cos x-3\sin x}{3\cos x+\sin x}\,dx=\log|3\cos x+\sin x|.$$ Therefore, $$I=\frac12\log|3\cos x+\sin x|+\frac{x}{2}+C.$$ 5. Compare with the given form: $$I=\frac12\left(\alpha x+\log|\beta\sin x+\gamma\cos x|\right)+C.$$ So we identify $$\alpha=1,\qquad \beta=1,\qquad \gamma=3.$$ 6. Compute the required value: $$\alpha+\frac{\gamma}{\beta}=1+\frac31=4.$$ 7. Option check: - A: $3$ ❌ - B: $7$ ❌ - C: $1$ ❌ - D: $4$ ✅ Therefore, the correct answer is **D**.

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View written solutionFree

Correct answer: D

We need to evaluate $I=\int \frac{2-\tan x}{3+\tan x}\,dx$ and compare it with $\frac12\left(\alpha x+\log_e|\beta\sin x+\gamma\cos x|\right)+C.$
Rewrite the integrand in terms of $\sin x$ and $\cos x$ :

=\frac{2\cos x-\sin x}{3\cos x+\sin x}.$$ So, $$I=\int \frac{2\cos x-\sin x}{3\cos x+\sin x}\,dx.$$ 3. Express the numerator in terms of the derivative of the denominator. Let $$D=3\cos x+\sin x.$$ Then $$D'=-3\sin x+\cos x=\cos x-3\sin x.$$ We want constants $A,B$ such that $$2\cos x-\sin x=A(\cos x-3\sin x)+B(3\cos x+\sin x).$$ Comparing coefficients: $$A+3B=2,\qquad -3A+B=-1.$$ Solving, $$A=\frac{5}{10}=\frac12,\qquad B=\frac12.$$ Thus, $$2\cos x-\sin x=\frac12(\cos x-3\sin x)+\frac12(3\cos x+\sin x).$$ Hence, $$\frac{2\cos x-\sin x}{3\cos x+\sin x}=\frac12\frac{\cos x-3\sin x}{3\cos x+\sin x}+\frac12.$$ 4. Integrate: $$I=\frac12\int \frac{\cos x-3\sin x}{3\cos x+\sin x}\,dx+\frac12\int dx.$$ Since $\cos x-3\sin x=(3\cos x+\sin x)'$, $$\int \frac{\cos x-3\sin x}{3\cos x+\sin x}\,dx=\log|3\cos x+\sin x|.$$ Therefore, $$I=\frac12\log|3\cos x+\sin x|+\frac{x}{2}+C.$$ 5. Compare with the given form: $$I=\frac12\left(\alpha x+\log|\beta\sin x+\gamma\cos x|\right)+C.$$ So we identify $$\alpha=1,\qquad \beta=1,\qquad \gamma=3.$$ 6. Compute the required value: $$\alpha+\frac{\gamma}{\beta}=1+\frac31=4.$$ 7. Option check: - A: $3$ ❌ - B: $7$ ❌ - C: $1$ ❌ - D: $4$ ✅ Therefore, the correct answer is **D**.