Question

If \[ \int \frac{2\cos x+3\sin x} {3\cos x+4\sin x}\,dx = Ax+B\log|3\cos x+4\sin x|+C, \] then \[ A\cdot B= \]

• A. \[ -\frac{18}{625} \]
• B. \[ \frac{18}{625} \]
• C. \[ -\frac{18}{25} \]
• D. \[ \frac{18}{25} \]

Hint:

For integrals of the form \[ \int \frac{a\cos x+b\sin x}{c\cos x+d\sin x}\,dx, \] write the numerator as \[ A(c\cos x+d\sin x)+B(-c\sin x+d\cos x), \] because \[ \frac{d}{dx}(c\cos x+d\sin x) = -c\sin x+d\cos x. \]

Answer 1

The Correct Option is A

Concept: Write numerator as linear combination of denominator and its derivative: \(2\cos x+3\sin x = A(3\cos x+4\sin x) + B(-3\sin x+4\cos x)\). Integrate, extract A and B, then multiply.

Step 1: Equate coefficients: \(\cos x: 3A+4B=2\); \(\sin x: 4A-3B=3\). Solve: multiply first by 3: \(9A+12B=6\); second by 4: \(16A-12B=12\). Add: \(25A=18 \Rightarrow A=18/25\). Then \(3(18/25)+4B=2 \Rightarrow 54/25+4B=2 \Rightarrow 4B = -4/25 \Rightarrow B = -1/25\).

Step 2: Integral: \(\int \frac{2\cos x+3\sin x}{3\cos x+4\sin x}dx = A x + B\log|3\cos x+4\sin x| + C\).

Step 3: \(A \cdot B = \frac{18}{25} \cdot \left(-\frac{1}{25}\right) = -\frac{18}{625}\).

Step 4: Write the final answer. \(\boxed{-\frac{18}{625}}\)