Question

If \(\quad \int \frac{3 \sin x \cos x}{4 \sin x+7} \mathrm{~d} x=\mathrm{A} \sin x-\mathrm{Blog}(4 \sin x+7)+\mathrm{c}\) where c is the constant of integration, then the value of \(\mathrm{A}+\mathrm{B}\) is equal to

• A. \(\frac{9}{16}\)
• B. \(\frac{-9}{16}\)
• C. \(\frac{33}{16}\)
• D. \(\frac{-33}{16}\)

Hint:

Put t equal to sin x, divide the numerator by the linear denominator and compare with the given form.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Let \(t=\sin x\), so \(dt=\cos x\,dx\). The integral becomes \(3\int\dfrac{t}{4t+7}dt\).

Step 2: Calculation
Write \(\dfrac{t}{4t+7}=\dfrac{1}{4}\left(1-\dfrac{7}{4t+7}\right)\).
So the integral is \(\dfrac{3}{4}\left[t-\dfrac{7}{4}\log(4t+7)\right]+c\).

Step 3: Compare
\(=\dfrac{3}{4}\sin x-\dfrac{21}{16}\log(4\sin x+7)+c\).
So \(A=\dfrac{3}{4}\) and \(B=\dfrac{21}{16}\).

Step 4: Add
\(A+B=\dfrac{12}{16}+\dfrac{21}{16}=\dfrac{33}{16}\), option (C).

Final Answer:
The value of \(A+B\) is \(\dfrac{33}{16}\). \[ \boxed{\text{(C) }\dfrac{33}{16}} \]