Question

The value of \(\int sin4xcos3x\,dx\) is

• A. \(-\frac{1}{14}cos7x-\frac{1}{2}cosx+c\)
• B. \(-\frac{1}{14}cos7x+\frac{1}{2}cosx+c\)
• C. \(\frac{1}{14}cos7x-\frac{1}{2}cosx+c\)
• D. \(\frac{1}{14}cos7x+\frac{1}{2}cosx+c\)

Hint:

Convert the product to a sum using 2 sin A cos B.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Use the product-to-sum identity \(2\sin A\cos B = \sin(A+B) + \sin(A-B)\).

Step 2: Apply:
\[ \sin 4x\cos 3x = \frac12\left[\sin 7x + \sin x\right] \]

Step 3: Integrate:
\[ \int \frac12(\sin 7x + \sin x)\,dx = \frac12\left(-\frac{\cos 7x}{7} - \cos x\right) + c = -\frac{1}{14}\cos 7x - \frac12\cos x + c \]

Step 4: Why the other options are wrong.
Options with \(+\frac1{14}\cos 7x\) or \(+\frac12\cos x\) have sign errors, because the integral of \(\sin\) is \(-\cos\).

Final Answer:
The integral is \(-\frac{1}{14}\cos 7x - \frac12\cos x + c\), option (A). \[ \boxed{-\frac{1}{14}\cos 7x-\frac{1}{2}\cos x+c} \]