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: Check by differentiation:
Differentiate $-\frac1{14}\cos7x - \frac12\cos x$: $\frac{7}{14}\sin 7x + \frac12\sin x = \frac12(\sin 7x + \sin x)$.

Step 2: Compare:
$\frac12(\sin 7x + \sin x) = \sin 4x\cos 3x$ by the identity, so option (A) matches.

Step 3: Signs:
The minus signs come from $\int\sin kx\,dx = -\frac{\cos kx}{k}$.

Final Answer:
Option (A). \[ \boxed{-\frac{1}{14}\cos 7x-\frac{1}{2}\cos x+c} \]