How do you integrate #inte^(3x)cos^2xdx# using integration by parts?

Question

Luke Alvarez · Accepted Answer

#inte^(3x)cos^2xdx=e^{3x}/78(9cos(2x)+6sin(2x)+13)#

This integral is easily solved by applying Moivre's identity's consequence.

#(e^{ix}+e^{-ix})/2 = cos(x)# #(e^{ix}-e^{-ix})/(2i) = sin(x)#

because

#cos^2x equiv (e^{2ix} +2+e^{-2ix})/4#

then

#inte^(3x)cos^2xdx equiv 1/4int (e^{3x+2ix} +2e^{3x}+e^{3x-2ix})dx#

#=e^{3x}/4((e^{2ix})/(3+2i)+2/3+e^{-2ix}/(3-2i))# #=e^{3x}/4((3-2i)/13e^{2ix}+2/3+(3+2i)/13e^{-2ix})# #=e^{3x}(3/26(e^{2ix}+e^{-2ix})/2+1/13(e^{2ix}-e^{-2ix})/(2i)+1/6)# #=e^{3x}/78(9cos(2x)+6sin(2x)+13)#

Paisley Johnson · Answer

undefined To integrate ∫e^(3x)cos^2(x) dx using integration by parts, follow these steps:

1. Choose u and dv:
   Let u = cos^2(x) and dv = e^(3x) dx.

2. Find du and v:
   Differentiate u to find du, and integrate dv to find v.

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

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

5. Evaluate the resulting integral.

Here are the steps in more detail:

1. Choose u and dv:
   u = cos^2(x)
   dv = e^(3x) dx

2. Find du and v:
   Differentiate u to find du:
   du = -2cos(x)sin(x) dx
   Integrate dv to find v:
   v = (1/3)e^(3x)

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

4. Substitute the values of u, dv, du, and v into the formula:
   ∫e^(3x)cos^2(x) dx = (1/3)e^(3x)cos^2(x) - ∫(1/3)e^(3x)(-2cos(x)sin(x)) dx

5. Evaluate the resulting integral.

That's the integration by parts method for integrating the given function.