To evaluate the integral

$\int_0^{\infty} \sin(qx) \left(0.191 e^{-0.279x/a} + 0.474 e^{-0.637x/a} + 0.355 e^{-1.919x/a}\right) \, dx$

we can proceed by splitting it into three separate integrals, as follows:

$\int_0^{\infty} \sin(qx) \left(0.191 e^{-0.279x/a} + 0.474 e^{-0.637x/a} + 0.355 e^{-1.919x/a}\right) \, dx = 0.191 \int_0^{\infty} \sin(qx) e^{-0.279x/a} \, dx + 0.474 \int_0^{\infty} \sin(qx) e^{-0.637x/a} \, dx + 0.355 \int_0^{\infty} \sin(qx) e^{-1.919x/a} \, dx.$

Each of these integrals has the general form:

$\int_0^{\infty} \sin(qx) e^{-\alpha x} \, dx,$

where $\alpha$ is a constant. The result of this integral is known and can be evaluated as:

$\int_0^{\infty} \sin(qx) e^{-\alpha x} \, dx = \frac{q}{\alpha^2 + q^2}.$

We apply this result to each term individually:

1. For $0.191 \int_0^{\infty} \sin(qx) e^{-0.279x/a} \, dx$: $= 0.191 \cdot \frac{q}{(0.279/a)^2 + q^2} = \frac{0.191q}{0.279^2/a^2 + q^2}.$

2. For $0.474 \int_0^{\infty} \sin(qx) e^{-0.637x/a} \, dx$: $= 0.474 \cdot \frac{q}{(0.637/a)^2 + q^2} = \frac{0.474q}{0.637^2/a^2 + q^2}.$

3. For $0.355 \int_0^{\infty} \sin(qx) e^{-1.919x/a} \, dx$: $= 0.355 \cdot \frac{q}{(1.919/a)^2 + q^2} = \frac{0.355q}{1.919^2/a^2 + q^2}.$

Finally, summing these three results, we obtain the solution:

$\int_0^{\infty} \sin(qx) \left(0.191 e^{-0.279x/a} + 0.474 e^{-0.637x/a} + 0.355 e^{-1.919x/a}\right) \, dx = \frac{0.191q}{0.279^2/a^2 + q^2} + \frac{0.474q}{0.637^2/a^2 + q^2} + \frac{0.355q}{1.919^2/a^2 + q^2}.$

Let me know if you would like further details or have any questions about the solution.

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Here are 5 additional questions you might explore based on this topic:

1. How does the solution change if we replace the sine function with a cosine?
2. What happens if the integral limits are changed to $[0, b]$ instead of $[0, \infty]$?
3. How would the result change if different exponential terms were used?
4. Can this approach be generalized to solve similar integrals with other functions?
5. What is the effect of changing the constant $a$ on the value of the integral?

Tip: When solving integrals with exponential and trigonometric functions, breaking down the integral into smaller parts can simplify the evaluation significantly.