Question

\( \int \frac{2\sin x - 3\cos x}{4\cos x - 3\sin x} dx = \)

• A. \( \frac{1}{25}[17\log|4\cos x-3\sin x|-6x]+c \)
• B. \( \frac{1}{25}[x-18\log|4\cos x-3\sin x|]+c \)
• C. \( \frac{1}{25}[\log|4\cos x-3\sin x|-18x]+c \)
• D. \( \frac{1}{25}[17x-6\log|4\cos x-3\sin x|]+c \)

Hint:

Solve the system of linear equations carefully. \( A \) is the coefficient of the linear term \( x \), and \( B \) is the coefficient of the logarithmic term.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

To integrate a rational trigonometric function of the form \( \frac{a \sin x + b \cos x}{c \sin x + d \cos x} \), we express the numerator as a linear combination of the denominator and its derivative.
Step 2: Key Formula or Approach:

Write \( \text{Numerator} = A(\text{Denominator}) + B(\frac{d}{dx}\text{Denominator}) \). Then \( \int \frac{Nr}{Dr} dx = \int (A + B \frac{Dr'}{Dr}) dx = Ax + B \ln|Dr| + C \).
Step 3: Detailed Explanation:

Let \( 2\sin x - 3\cos x = A(4\cos x - 3\sin x) + B(-4\sin x - 3\cos x) \). Equate coefficients of \( \sin x \) and \( \cos x \): For \( \sin x \): \( 2 = -3A - 4B \) \quad ...(i) For \( \cos x \): \( -3 = 4A - 3B \) \quad ...(ii) Multiply (i) by 4 and (ii) by 3: \( 8 = -12A - 16B \) \( -9 = 12A - 9B \) Add them: \( -1 = -25B \implies B = \frac{1}{25} \). Substitute \( B \) into (i): \( 2 = -3A - 4(1/25) \) \( 2 + \frac{4}{25} = -3A \) \( \frac{54}{25} = -3A \implies A = -\frac{18}{25} \). The integral becomes: \[ \int \left( -\frac{18}{25} + \frac{1}{25} \frac{Dr'}{Dr} \right) dx \] \[ = -\frac{18}{25}x + \frac{1}{25} \ln|4\cos x - 3\sin x| + c \] \[ = \frac{1}{25} \left[ \ln|4\cos x - 3\sin x| - 18x \right] + c \]
Step 4: Final Answer:

Matches Option (C).