How do I evaluate #inte^(2x)cosx dx# by parts?

Question

William Abdalla · Accepted Answer

I´ll start by integrating #e^(2x)# and leaving #cos(x)# as it is and then derive it leaving the integrated part as it is. #e^(2x)/2*cos(x)-inte^(2x)/2*(-sin(x))dx=#
#=e^(2x)/2*cos(x)+1/2inte^(2x)*(sin(x))dx=# by parts again:
#=e^(2x)/2*cos(x)+1/2[e^(2x)/2*(sin(x))-inte^(2x)/2*cos(x)dx]=#
#=e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))-1/4inte^(2x)*cos(x)dx# So your integral is:
#inte^(2x)cos(x)dx=e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))-1/4inte^(2x)*cos(x)dx# Now I can take to the left the integral: #-1/4inte^(2x)*cos(x)dx#
Giving: #inte^(2x)cos(x)dx+1/4inte^(2x)*cos(x)dx=e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))# #5/4inte^(2x)*cos(x)dx=e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))# and finally:
#inte^(2x)*cos(x)dx=4/5[e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))]+c#

Isabella Ackerman · Answer

undefined To evaluate the integral ∫e^(2x)cos(x) dx by parts:

1. Choose u and dv such that:
   - u is differentiable and can be easily integrated.
   - dv can be differentiated easily.

2. Calculate du and v:
   - du is the derivative of u with respect to x.
   - v is the integral of dv with respect to x.

3. Apply the integration by parts formula:
   ∫u dv = uv - ∫v du

4. Substitute the values of u, dv, du, and v into the integration by parts formula.

5. Evaluate the resulting integral and simplify if possible.