Question 12

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Evaluate: $\int (\sin \pi x-3\sin 3x)dx$

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Solution

We use the standard formula for integrating sine with a linear argument:

$\int \sin (ax)dx=-\frac{\cos (ax)}{a}+C$

Applying the linearity property of integration:

$\int (\sin \pi x-3\sin 3x)dx=\int \sin \pi xdx-3\int \sin 3xdx$

First term: $a=\pi$

$\int \sin \pi xdx=-\frac{\cos \pi x}{\pi }$

Second term: $a=3$

$3\int \sin 3xdx=3\cdot \left(-\frac{\cos 3x}{3}\right)=-\cos 3x$

Combining both terms:

$\int (\sin \pi x-3\sin 3x)dx=-\frac{\cos \pi x}{\pi }-(-\cos 3x)+C$

$\boxed{=-\frac{1}{\pi }\cos \pi x+\cos 3x+C}$