In my calculus class, we're working on integration by parts specifically, but the worksheet I was given can utilize various methods. I've been toying with these for a while, but I don't think I did them correctly. Can someone please help me? I'm stuck on a couple of them.
The indefinite integral of (x^3)(e^x) dx
The indefinite integral of [e^(2x)]sin3x dx
Thank you so much! These exponentials are driving me crazy! ><
The indefinite integral of (x^3)(e^x) dx
The indefinite integral of [e^(2x)]sin3x dx
Thank you so much! These exponentials are driving me crazy! ><
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Both of these involve Integration by Parts.
∫ x^3 e^x dx
Let u = x^3 and dv = e^x dx
then du = 3x^2 dx and v = e^x
x^3 e^x - ∫ 3x^2 e^x dx
U = x^2 and dV = e^x dx
dU = 2x dx and V = e^x
x^3 e^x - 3x^2 e^x + ∫ 2x e^x dx
W = x and dT = e^x dx
dW = dx and T = e^x
x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x + C
***********************
∫ e^(2x) sin 3x dx
Let u = e^(2x) and dv = sin 3x dx
then du = 2e^(2x) dx and v = -1/3 cos 3x
-1/3 e^(2x) cos 3x + 2/3 ∫ e^(2x) cos 3x dx
Let U = e^(2x) and dV = cos 3x dx
then dU = 2 e^(2x) dx and V = 1/3 sin 3x
∫ e^(2x) sin 3x dx = -1/3 e^(2x) cos 3x + 2/9 e^(2x) sin 3x - 4/9 ∫ e^(2x) sin 3x dx
13/9 ∫ e^(2x) sin 3x dx = -1/3 e^(2x) cos 3x + 2/9 e^(2x) sin 3x
So, ∫ e^(2x) sin 3x dx = -3/13 e^(2x) cos 3x + 2/13 e^(2x) sin 3x + C
∫ x^3 e^x dx
Let u = x^3 and dv = e^x dx
then du = 3x^2 dx and v = e^x
x^3 e^x - ∫ 3x^2 e^x dx
U = x^2 and dV = e^x dx
dU = 2x dx and V = e^x
x^3 e^x - 3x^2 e^x + ∫ 2x e^x dx
W = x and dT = e^x dx
dW = dx and T = e^x
x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x + C
***********************
∫ e^(2x) sin 3x dx
Let u = e^(2x) and dv = sin 3x dx
then du = 2e^(2x) dx and v = -1/3 cos 3x
-1/3 e^(2x) cos 3x + 2/3 ∫ e^(2x) cos 3x dx
Let U = e^(2x) and dV = cos 3x dx
then dU = 2 e^(2x) dx and V = 1/3 sin 3x
∫ e^(2x) sin 3x dx = -1/3 e^(2x) cos 3x + 2/9 e^(2x) sin 3x - 4/9 ∫ e^(2x) sin 3x dx
13/9 ∫ e^(2x) sin 3x dx = -1/3 e^(2x) cos 3x + 2/9 e^(2x) sin 3x
So, ∫ e^(2x) sin 3x dx = -3/13 e^(2x) cos 3x + 2/13 e^(2x) sin 3x + C