Evaluate the integral: I = integral (sin 5x - sin 5
Evaluate the integral: $$ I = \int (\sin 5x - \sin 5\alpha)\,dx, $$ where $\alpha$ is a constant.
Explanation of Method
Since $\alpha$ is a fixed parameter (not the variable of integration), $\sin 5\alpha$ is just a number. Integrating it with respect to $x$ therefore gives $\sin(5\alpha)\cdot x$, the same as integrating any other constant, while the $\sin 5x$ term is integrated in the usual way.
- Split the integral into two terms using linearity.
- Integrate the $\sin 5x$ term using the standard sine antiderivative.
- Integrate the constant term $\sin 5\alpha$ with respect to $x$.
- Combine the results and add the constant of integration.
Step 1 · Split the integral
We evaluate the given integral by integrating each term separately:
$$ I = \int \sin 5x\,dx - \int \sin 5\alpha\,dx. $$
Since $\alpha$ is a constant, $\sin 5\alpha$ is also a constant. We integrate each term separately:
$$ \int \sin 5x\,dx = -\frac{1}{5}\cos 5x, $$
$$ \int \sin 5\alpha\,dx = \sin 5\alpha \int dx = \sin 5\alpha \cdot x. $$
Step 2 · Combine the results
Thus, the solution to the integral is:
$$ I = -\frac{1}{5}\cos 5x - x\sin 5\alpha + C, $$
where $C$ is the constant of integration.
Summary Table
| Term | Antiderivative |
|---|---|
| $\sin 5x$ | $-\tfrac15\cos 5x$ |
| $-\sin 5\alpha$ (constant) | $-x\sin 5\alpha$ |
Additional Notes
- The integrand is defined for all real $x$, since $\sin$ is defined everywhere.
- Recognizing that $\alpha$ is a parameter, not the integration variable, is the key idea here — $\sin 5\alpha$ behaves exactly like any other numeric constant (such as $7$ or $\pi$) when integrating with respect to $x$.
- The result can be checked by differentiating $-\tfrac15\cos5x - x\sin5\alpha$ with respect to $x$ and confirming it returns $\sin5x - \sin5\alpha$.
Result
The solution to the integral is: