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: Express the numerator as \[ 2\cos x+3\sin x = A(3\cos x+4\sin x) + B(-3\sin x+4\cos x), \] so that the second term becomes the derivative of the denominator.

Step 1: Compare coefficients. We require \[ 2\cos x+3\sin x = A(3\cos x+4\sin x) + B(-3\sin x+4\cos x). \] Comparing coefficients of \(\cos x\) and \(\sin x\), \[ 3A+4B=2, \] \[ 4A-3B=3. \] \[ \cdots (1) \]

Step 2: Solve for \(A\) and \(B\). Multiplying the first equation by \(3\), \[ 9A+12B=6. \] Multiplying the second equation by \(4\), \[ 16A-12B=12. \] Adding, \[ 25A=18. \] \[ A=\frac{18}{25}. \] Substituting into \[ 3A+4B=2, \] \[ \frac{54}{25}+4B=2. \] \[ 4B=-\frac4{25}. \] \[ B=-\frac1{25}. \]

Step 3: Integrate. Hence \[ \frac{2\cos x+3\sin x} {3\cos x+4\sin x} = \frac{18}{25} -\frac1{25} \frac{-3\sin x+4\cos x} {3\cos x+4\sin x}. \] Therefore \[ \int \frac{2\cos x+3\sin x} {3\cos x+4\sin x}\,dx = \frac{18}{25}x -\frac1{25} \log|3\cos x+4\sin x| +C. \] Thus \[ A=\frac{18}{25}, \qquad B=-\frac1{25}. \]

Step 4: Compute \(A\cdot B\). \[ A\cdot B = \frac{18}{25} \left(-\frac1{25}\right). \] \[ = -\frac{18}{625}. \]

Step 5: Write the final answer. \[ \boxed{-\frac{18}{625}} \]