1.evaluate (integral sign)x cos 3x dx
A.1/6 x^2 sin 3x + C
B.1/3 x sin 3x -1/2 sin 3x +C
C.1/3 x sin 3x +1/9 cos 3x +C << my choice.
D. 1/2 x^2 +1/18 sin^2 3x +C
2.evaluate (integral sign)xe(power of 2x)dx
A.1/6 x^2 e(to the power of 3x)+C
B.1/2 xe(to the power of 2x)-1/2 e(to
the power of 2x)+C
C.1/2xe(to the power of 2x) -1/4 e(to the power of 2x) +C
D.1/2 x^2 -1/8 e(to the power of 4x +C
I don't know the answer to this one.
1. C is Correct
Integration by parts,
∫xcos(3x)dx
=(1/3)xsin(3x)-(1/3)∫cos(3x)dx
= ....
2. Same as 1. Use integration by parts
∫xe2xdx
=(1/2)xe2x-(1/2)∫e2xdx
=...
Perfect!
To solve the given integrals, we will use integration by parts. The formula for integration by parts is ∫(u dv) = uv - ∫(v du), where u and v are differentiable functions.
1. ∫x cos(3x) dx:
Let u = x and dv = cos(3x) dx.
Taking the derivative of u, we get du = dx.
Integrating dv, we get v = (1/3) sin(3x).
Using the integration by parts formula:
∫x cos(3x) dx = (1/3) x sin(3x) - ∫(1/3) sin(3x) dx
To integrate ∫sin(3x) dx, we can use the substitution method:
Let u = 3x, then du = 3 dx.
∫sin(3x) dx = (1/3) ∫sin(u) du = -(1/3) cos(u) + C = -(1/3) cos(3x) + C
Substituting this value back into the previous equation:
∫x cos(3x) dx = (1/3) x sin(3x) + (1/9) cos(3x) + C
Therefore, the correct option is C. 1/3 x sin(3x) + 1/9 cos(3x) + C.
2. ∫x e^(2x) dx:
Here, we choose u = x and dv = e^(2x) dx.
Taking the derivative of u, we get du = dx.
Integrating dv using the power rule for integration, we get v = (1/2) e^(2x).
Using the integration by parts formula:
∫x e^(2x) dx = (1/2) x e^(2x) - ∫(1/2) e^(2x) dx
To integrate ∫e^(2x) dx, we can simply apply the power rule for integration:
∫e^(2x) dx = (1/2) e^(2x) + C
Substituting this value back into the previous equation:
∫x e^(2x) dx = (1/2) x e^(2x) - (1/4) e^(2x) + C
Therefore, the correct option is C. 1/2x e^(2x) - 1/4 e^(2x) + C.
I hope this helps! Let me know if you have any further questions.