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Set up the integral for the given expression.

Mathematics

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Answer

−13cos⁡(3x)−13e3x+C-\frac{1}{3}\cos(3x) - \frac{1}{3}e^{3x} + C

Step-by-step solution

3 steps
  1. Set up the integral for the given expression.

    We need to integrate sin⁡(3x)−e3x\sin(3x) - e^{3x} with respect to xx.

    ∫(sin⁡(3x)−e3x) dx\int (\sin(3x) - e^{3x})\,dx
  2. Integrate each term separately.

    The integral of a difference is the difference of the integrals.

    ∫sin⁡(3x) dx−∫e3x dx\int \sin(3x)\,dx - \int e^{3x}\,dx

    For the first term, ∫sin⁡(ax) dx=−1acos⁡(ax)+C\int \sin(ax)\,dx = -\frac{1}{a}\cos(ax) + C. Here a=3a=3.

    ∫sin⁡(3x) dx=−13cos⁡(3x)\int \sin(3x)\,dx = -\frac{1}{3}\cos(3x)

    For the second term, ∫eax dx=1aeax+C\int e^{ax}\,dx = \frac{1}{a}e^{ax} + C. Here a=3a=3.

    ∫e3x dx=13e3x\int e^{3x}\,dx = \frac{1}{3}e^{3x}
  3. Combine the results and add the constant of integration.

    −13cos⁡(3x)−13e3x+C-\frac{1}{3}\cos(3x) - \frac{1}{3}e^{3x} + C
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