To evaluate the integral ∫e3xcos5xdx, we will use integration by parts twice. The formula for integration by parts is ∫udv=uv−∫vdu.
Let I=∫e3xcos5xdx.
Step 1: Apply integration by parts for the first time.
Let u=cos5x and dv=e3xdx.
Then, we find du and v:
du=−5sin5xdx
v=∫e3xdx=31e3x
Substitute these into the integration by parts formula:
I=(cos5x)(31e3x)−∫(31e3x)(−5sin5x)dx
I=31e3xcos5x+35∫e3xsin5xdx
Step 2: Apply integration by parts for the second time to the new integral ∫e3xsin5xdx.
Let I1=∫e3xsin5xdx.
Let u=sin5x and dv=e3xdx.
Then, we find du and v:
du=5cos5xdx
v=∫e3xdx=31e3x
Substitute these into the integration by parts formula:
I1=(sin5x)(31e3x)−∫(31e3x)(5cos5x)dx
I1=31e3xsin5x−35∫e3xcos5xdx
Notice that the integral on the right side is our original integral I.
So, we have:
I1=31e3xsin5x−35I
Step 3: Substitute I1 back into the equation for I from Step 1.
I=31e3xcos5x+35(31e3xsin5x−35I)
I=31e3xcos5x+95e3xsin5x−925I
Step 4: Solve the equation for I.
Add 925I to both sides:
I+925I=31e3xcos5x+95e3xsin5x
(1+925)I=31e3xcos5x+95e3xsin5x
(99+25)I=31e3xcos5x+95e3xsin5x
934I=31e3xcos5x+95e3xsin5x
Multiply both sides by 349:
I=349(31e3xcos5x+95e3xsin5x)
I=349⋅31e3xcos5x+349⋅95e3xsin5x
I=343e3xcos5x+345e3xsin5x
Step 5: Add the constant of integration.
I=34e3x(3cos5x+5sin5x)+C
The final answer is 34e3x(3cos5x+5sin5x)+C.