Calculus

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)

asked by Anonymous
  1. 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.

    posted by drwls
  2. 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)

    posted by Anonymous

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