Evaluate the integral of e^(2x) * cos(3x) dx.

Answer: 113e2x(2cos⁡(3x)+3sin⁡(3x))+C\frac{1}{13}e^{2x}(2\cos(3x) + 3\sin(3x)) + C

Step-by-step solution

Step 1: Apply Integration by Parts

We will use integration by parts to solve this integral. Let u=cos⁡(3x)u = \cos(3x) and dv=e2x dxdv = e^{2x} \, dx. We then find dudu and v.

Step 2: Calculate dudu and vv

Differentiating u=cos⁡(3x)u = \cos(3x) gives du=−3sin⁡(3x) dxdu = -3\sin(3x) \, dx. Integrating dv=e2x dxdv = e^{2x} \, dx gives v=12e2xv = \frac{1}{2}e^{2x}.

Step 3: Substitute into Integration by Parts Formula

Substitute the calculated values of u,v,du,dvu, v, du, dv into the integration by parts formula. This results in a new integral, ∫e2xsin⁡(3x) dx\int e^{2x}\sin(3x) \, dx, which also requires integration by parts.

Step 4: Apply Integration by Parts Again

For the new integral, we again apply integration by parts. Let u′=sin⁡(3x)u' = \sin(3x) and dv′=e2x dxdv' = e^{2x} \, dx. We find du′=3cos⁡(3x) dxdu' = 3\cos(3x) \, dx and v′=12e2xv' = \frac{1}{2}e^{2x}. Substituting these into the formula gives an expression that includes the original integral.

Step 5: Solve for the Original Integral

Let I=∫e2xcos⁡(3x) dxI = \int e^{2x} \cos(3x) \, dx. Substitute the result from the second integration by parts back into the equation from the first integration by parts. Then, rearrange the equation to solve for II. Finally, add the constant of integration C.

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