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Evaluate the integral integral e^x x dx

Mathematics

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Answer

ex(sin⁡x−cos⁡x)2+C\frac{e^x (\sin x - \cos x)}{2} + C

Step-by-step solution

5 steps

We need to evaluate the integral:

∫exsin⁡x dx\int e^x \sin x \, dx

This integral requires integration by parts twice. The formula for integration by parts is ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du.

  1. Apply integration by parts for the first time.

    Let I=∫exsin⁡x dxI = \int e^x \sin x \, dx.
    Choose u=sin⁡xu = \sin x and dv=ex dxdv = e^x \, dx.
    Then, du=cos⁡x dxdu = \cos x \, dx and v=exv = e^x.
    Applying the formula:

    I=exsin⁡x−∫excos⁡x dx(∗)I = e^x \sin x - \int e^x \cos x \, dx \quad (*)

  2. Apply integration by parts again to the new integral ∫excos⁡x dx\int e^x \cos x \, dx.

    Choose u=cos⁡xu = \cos x and dv=ex dxdv = e^x \, dx.
    Then, du=−sin⁡x dxdu = -\sin x \, dx and v=exv = e^x.
    Applying the formula:

    ∫excos⁡x dx=excos⁡x−∫ex(−sin⁡x) dx\int e^x \cos x \, dx = e^x \cos x - \int e^x (-\sin x) \, dx ∫excos⁡x dx=excos⁡x+∫exsin⁡x dx\int e^x \cos x \, dx = e^x \cos x + \int e^x \sin x \, dx

  3. Substitute the result from Step 2 back into equation (∗)(*) from Step 1.

    I=exsin⁡x−(excos⁡x+∫exsin⁡x dx)I = e^x \sin x - \left(e^x \cos x + \int e^x \sin x \, dx\right) I=exsin⁡x−excos⁡x−∫exsin⁡x dxI = e^x \sin x - e^x \cos x - \int e^x \sin x \, dx

  4. Notice that the original integral II appears on the right side.

    Move it to the left side to solve for II.

    I+∫exsin⁡x dx=exsin⁡x−excos⁡xI + \int e^x \sin x \, dx = e^x \sin x - e^x \cos x 2I=exsin⁡x−excos⁡x2I = e^x \sin x - e^x \cos x

  5. Solve for II and add the constant of integration.

    I=exsin⁡x−excos⁡x2+CI = \frac{e^x \sin x - e^x \cos x}{2} + C I=ex(sin⁡x−cos⁡x)2+CI = \frac{e^x (\sin x - \cos x)}{2} + C

    The integral is ex(sin⁡x−cos⁡x)2+C\frac{e^x (\sin x - \cos x)}{2} + C.

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