Question

The value of $\int \sin 8x\cos 3xdx$ is
(where $C$ is constant of integration)

• A. $\frac{1}{2}(-\frac{\cos 11x}{11}-\frac{\cos 5x}{5})+C$
• B. $\frac{1}{2}(-\frac{\cos 11x}{11}+\frac{\cos 5x}{5})+C$
• C. $\frac{1}{2}(\frac{\cos 11x}{11}-\frac{\cos 5x}{5})+C$
• D. $\frac{1}{2}(-\frac{\cos 11x}{11}+\frac{\cos 5x}{5})+C$

Answer 1

The correct option is A $\frac{1}{2}(-\frac{\cos 11x}{11}-\frac{\cos 5x}{5})+C$

Let $I=\int \sin 8x\cos 3xdx$
$I=\frac{1}{2}\int 2\sin 8x\cos 3xdx=\frac{1}{2}\int (\sin 11x+\sin 5x)dx$
$(\because 2\sin a\cos b=\sin (a+b)+\sin (a-b\left)\right)$

$\Rightarrow I=\frac{1}{2}(-\frac{\cos 11x}{11}-\frac{\cos 5x}{5})+C$