Question

If \(\int \frac{2\,\tan x+3}{3\,\tan x+4}dx=Ax+B\log |3\,\sin x+4\,\cos x|+c\), then the values of \(A\) and \(B\) respectively are

• A. \(-\frac{18}{25},\frac{1}{25}\)
• B. \(-\frac{18}{25},-\frac{1}{25}\)
• C. \(\frac{18}{25},-\frac{1}{25}\)
• D. \(\frac{18}{25},\frac{1}{25}\)

Hint:

Write in sin and cos, express the numerator as a combination of the denominator and its derivative.

Answer 1

The Correct Option is D

Step 1: Differentiate the Answer Form
$\dfrac{d}{dx}\left[Ax+B\log|3\sin x+4\cos x|\right]=A+B\dfrac{3\cos x-4\sin x}{3\sin x+4\cos x}$.

Step 2: Match with the Integrand
This equals $\dfrac{(3A-4B)\sin x+(4A+3B)\cos x}{3\sin x+4\cos x}$, which must equal $\dfrac{2\sin x+3\cos x}{3\sin x+4\cos x}$.

Step 3: Check Option (D)
For $A=\dfrac{18}{25}$, $B=\dfrac{1}{25}$: $3A-4B=\dfrac{54-4}{25}=2$ and $4A+3B=\dfrac{72+3}{25}=3$. Both match.

Final Answer:
Only A = 18/25 and B = 1/25 satisfy both coefficient conditions. \[ \boxed{\text{(D) }\dfrac{18}{25},\dfrac{1}{25}} \]