∫ e^x cos(x) dx using integration by parts

Sure, let's use the integration by parts formula, which is ∫udv = uv - ∫vdu. We need to choose u and dv such that the integral becomes simpler when we differentiate u and integrate dv.

Step 1: Choose u and dv Let's choose: u = e^x, dv = cos(x) dx

Step 2: Differentiate u and integrate dv This gives us: du = e^x dx, v = ∫cos(x) dx = sin(x)

Step 3: Substitute into the integration by parts formula Substituting these into the integration by parts formula gives: ∫e^x cos(x) dx = e^x sin(x) - ∫sin(x) e^x dx

Now, we have a new integral ∫sin(x) e^x dx, which we can solve again using integration by parts.

Step 4: Solve the new integral using integration by parts Choose: u = e^x, dv = sin(x) dx

Differentiate u and integrate dv to get: du = e^x dx, v = -cos(x)

Substitute into the integration by parts formula to get: ∫sin(x) e^x dx = -e^x cos(x) - ∫-cos(x) e^x dx

Step 5: Substitute back into the original equation Substituting this back into the equation from step 3 gives: ∫e^x cos(x) dx = e^x sin(x) - (-e^x cos(x) - ∫-cos(x) e^x dx) Simplifying this gives: ∫e^x cos(x) dx = e^x sin(x) + e^x cos(x) + ∫e^x cos(x) dx

Step 6: Solve for the original integral We can see that the original integral ∫e^x cos(x) dx appears on both sides of the equation. We can solve for it by subtracting ∫e^x cos(x) dx from both sides, which gives: 0 = e^x sin(x) + e^x cos(x) So, the solution to the original integral is: ∫e^x cos(x) dx = e^x sin(x) + e^x cos(x) + C where C is the constant of integration.