Question

integrate:
e^2xcos3xdx

Answer 1

Let u = e^2x and dv = cos(3x)dx. Then, du = 2e^2x dx and v = sin(3x)/3

The integral may be rewritten as follows

∫u dv = uv - ∫v du

∫ e^(2x) cos(3x) dx = e^(2x) sin(3x)/3 - (2/3)∫e^(2x) sin(3x) dx

This requires us to do integration by parts again

p = e^(2x), dp = 2e^(2x) dx, dq = sin(3x) dx, q = -cos(3x)/3

Now, ∫p dq = pq - ∫q dp

∫ e^(2x) cos(3x) dx = e^(2x) sin(3x)/3 - (2/3)[-e^(2x) cos(3x)/3 + (2/3)∫e^(2x) cos(3x) dx]

= e^(2x) sin(3x)/3 + (2/9)e^(2x) cos(3x) - (4/9)∫e^(2x) cos(3x) dx

You have 9/9 of the original integral on the LHS and -4/9 of it on the RHS. So add 4/9 to both sides

(13/9)∫ e^(2x) cos(3x) dx = e^(2x) sin(3x)/3 + (2/9)e^(2x) cos(3x)

Finally, multiply both sides by 9/13 and add the constant of integration to get the final answer

∫ e^(2x) cos(3x) dx = (3/13) e^(2x) sin(3x) + (2/13) e^(2x) cos(3x) + C

Answer 2

medal please

Answer 3

.thank you

Answer 4

can you please explain where did you get the sin(3x)/3 in this equation
∫ e^(2x) cos(3x) dx = e^(2x) sin(3x)/3 - (2/3)[-e^(2x) cos(3x)/3 + (2/3)∫e^(2x) cos(3x) dx]
thank you

Answer 5

@Tai

Answer 6

yes it's for tai.

Answer 7

ty

Answer 8

@tai..can you please explain where did you get the sin(3x)/3 in this equation ∫ e^(2x) cos(3x) dx = e^(2x) sin(3x)/3 - (2/3)[-e^(2x) cos(3x)/3 + (2/3)∫e^(2x) cos(3x) dx] thank you

Answer 9

c/o: sin(3x)/3, was the differential of cos3x