I don't know if I did these problems correctly. Can you check them?
Use Integration by parts to solve problems.
integral x^3(lnx)dx
u=lnx dv=x^3dx
du=1/x v=x^4/4
Answer:(x^3)(lnx)-(x^4/16)
integral xcosxdx
x cosx
1 sinx
0 -cosx
Answer: xcosx+cosx
integral e^2x(sinx)dx
u=e^2x dv=sinxdx
du=2e^2x v=-cosx
-e^x(cosx)+integral(2e^2x)(cosx)
Answer: I don't know. Help!
integral (x^2)(e^2x)
x^2 e^2x
2x (1/2)e^2x
2 (1/4)e^2x
0 (1/8)e^2x
Answer: (1/2)(x^2)(e^2x)-(1/2)(x)(e^2x)+(1/4)(e^2x)
integral x^3(lnx)dx
u=lnx dv=x^3dx
du=1/x v=x^4/4
Answer:(x^4*lnx/4)-(x^4/16)
integral x cosxdx
u =x dv = cosxdx
du = 1 v = sin x
Answer: x sin x + cosx
Integral e^2x(sinx)dx
u=e^2x dv=sinxdx
du = 2e^2x v=-cosx
Answer -cos x*e^-2x +Integral 2 cos x e^2x
Now use the same integration by parts trick one more time to get a term that contains sin x e^2x on the right. Since you already have the same term on the left side, with a different coefficient, you can move all the sin x e^2x terms to one side of the equation and solve for it.
Can you check if I did it correctly?
integral e^2x(sinx)dx
u=e^2x dv=sinxdx
du=2e^2x v=-cosx
-e^x(cosx)+integral(2e^2x)(cosx)
u=2e^2x dv=cosxdx
du=e^2x v=sinx
Answer: -e^2x(cosx)+(2e^2x)(sinx)-(e^x)(sinx)
To check whether you've done these integration by parts correctly, we'll go through each problem one by one.
1. integral x^3(lnx)dx
To use integration by parts, we choose u = lnx and dv = x^3dx. Taking the derivatives and antiderivatives of u and v, we have du = (1/x)dx and v = (x^4/4).
Using the integration by parts formula: ∫ u dv = uv - ∫ v du, we have:
integral x^3(lnx)dx = (x^4/4)(lnx) - ∫ (x^4/4)(1/x)dx
Simplifying the integral on the right side: ∫ (x^4/4)(1/x)dx = ∫ x^3dx = x^4/16
Therefore, the final answer is:
integral x^3(lnx)dx = (x^4/4)(lnx) - x^4/16
Your answer is correct!
2. integral xcosxdx
For this problem, we choose u = x and dv = cosxdx. Taking the derivatives and antiderivatives, we have du = dx and v = sinx.
Using the integration by parts formula: ∫ u dv = uv - ∫ v du, we have:
integral xcosxdx = xsinx - ∫ sinxdx
The integral of sinx is -cosx. Therefore, the final answer is:
integral xcosxdx = xsinx + cosx
Your answer is correct!
3. integral e^2x(sinx)dx
For this problem, we choose u = e^2x and dv = sinxdx. Taking the derivatives and antiderivatives, we have du = 2e^2xdx and v = -cosx.
Using the integration by parts formula: ∫ u dv = uv - ∫ v du, we have:
integral e^2x(sinx)dx = -e^2x(cosx) - ∫ (-cosx)(2e^2x)dx
Now we have another integration by parts to perform. Using the integration by parts formula again, we choose u = -cosx and dv = 2e^2xdx. Taking the derivatives and antiderivatives, we have du = sinx dx and v = e^2x.
Now we can substitute the values back into the formula:
∫ (-cosx)(2e^2x)dx = -2e^2x(cosx) - ∫ (e^2x)(sinx)dx
So, the integral becomes:
integral e^2x(sinx)dx = -e^2x(cosx) - (-2e^2x(cosx) - ∫ (e^2x)(sinx)dx)
Simplifying further, we have:
integral e^2x(sinx)dx = -e^2x(cosx) + 2e^2x(cosx) + integral e^2x(sinx)dx
Now, we can move the remaining integral to the other side:
2e^2x(cosx) = -e^2x(cosx) + integral e^2x(sinx)dx
Solving for integral e^2x(sinx)dx, we get:
integral e^2x(sinx)dx = 2e^2x(cosx) + e^2x(cosx)
Your initial answer was incorrect. The correct answer is:
integral e^2x(sinx)dx = 2e^2x(cosx) + e^2x(cosx)
4. integral (x^2)(e^2x)
For this problem, we choose u = x^2 and dv = e^2xdx. Taking the derivatives and antiderivatives, we have du = 2x dx and v = (1/2)e^2x.
Using the integration by parts formula: ∫ u dv = uv - ∫ v du, we have:
integral (x^2)(e^2x) = (x^2)((1/2)e^2x) - ∫ ((1/2)e^2x)(2x)dx
Simplifying the integral on the right side: ∫ ((1/2)e^2x)(2x)dx = ∫ e^2xx dx = (1/4)e^2x(x^2) - (1/4)e^2x(x) + (1/8)e^2x
Therefore, the final answer is:
integral (x^2)(e^2x) = (1/2)(x^2)(e^2x) - (1/2)(x)(e^2x) + (1/4)(e^2x)
Your answer is correct!