Question Statement

Evaluate the integral:

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

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Background and Explanation

This problem requires integrating trigonometric functions with linear arguments (of the form $\sin (ax)$). The key prerequisite is knowing the standard integral formula for sine functions and applying the linearity property of integrals to handle the difference and constant coefficient.

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Solution

We evaluate the integral using the linearity property and the standard integration formula for sine functions.

First, split the integral into two separate integrals using the linearity property $\int [f(x)-g(x)]dx=\int f(x)dx-\int g(x)dx$:

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

Apply the standard integration formula $\int \sin (ax)dx=-\frac{\cos (ax)}{a}+C$ to each term:

For the first term with $a=\pi$: $\int \sin \pi xdx=\frac{-\cos \pi x}{\pi }$

For the second term with $a=3$: $\int \sin 3xdx=\frac{-\cos 3x}{3}$

Substituting these back and including the constant of integration $c$:

$\begin{array}{rl}I& =\frac{-\cos \pi x}{\pi }-3\left(\frac{-\cos 3x}{3}\right)+c\\ & =\frac{-\cos \pi x}{\pi }+\frac{3\cos 3x}{3}+c\\ & =-\frac{1}{\pi }\cos \pi x+\cos 3x+c\end{array}$

Verification by Differentiation

To confirm our answer is correct, we differentiate the result:

$\begin{array}{rl}\frac{d}{dx}\left(-\frac{1}{\pi }\cos \pi x+\cos 3x+c\right)& =\frac{d}{dx}\left(-\frac{1}{\pi }\cos \pi x\right)+\frac{d}{dx}(\cos 3x)+\frac{d}{dx}(c)\\ & =-\frac{1}{\pi }\cdot (-\sin \pi x)\cdot \pi +(-\sin 3x)\cdot 3+0\\ & =\sin \pi x-3\sin 3x\end{array}$

This matches the original integrand, confirming that:

$I=-\frac{1}{\pi }\cos \pi x+\cos 3x+c$

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Key Formulas or Methods Used

• Linearity of Integration: $\int [f(x)-g(x)]dx=\int f(x)dx-\int g(x)dx$ and $\int k\cdot f(x)dx=k\int f(x)dx$
• Standard Sine Integral: $\int \sin (ax)dx=-\frac{\cos (ax)}{a}+C$
• Chain Rule for Differentiation: $\frac{d}{dx}[\cos (ax)]=-a\sin (ax)$ (used for verification)

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Summary of Steps

1. Separate the integral using linearity: $\int \sin \pi xdx-3\int \sin 3xdx$
2. Apply the sine integral formula to each term: $-\frac{\cos \pi x}{\pi }-3\left(\frac{-\cos 3x}{3}\right)$
3. Simplify by canceling the 3s in the second term: $-\frac{1}{\pi }\cos \pi x+\cos 3x$
4. Add the constant of integration: $+c$
5. Verify by differentiating the result to recover the original integrand