Posted by **Anonymous** on Sunday, November 18, 2007 at 12:02am.

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)

- Calculus -
**drwls**, Sunday, November 18, 2007 at 12:20am
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.

- Calculus -
**Anonymous**, Sunday, November 18, 2007 at 12:46pm
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)

## Answer this Question

## Related Questions

- calc - integral of 1 to e^4 dx/x(1+lnx) 1+lnx = a => dx/x = da dx/ x(1+lnx...
- calculus - USE INTEGRATION BY PARTS TO FIND EACH INTEGRAL ƪx lnx dx
- CALCULUS - USE INTEGRATION BY PARTS TO FIND EACH INTEGRAL ƪx^5 Lnx dx
- calculus - how do you determine the convergence of : definite integral from 1--&...
- calculus - how do you solve the integral of 1/[(square root of x)(lnx)] from 2 ...
- Calc 2 - a. Integral (x^2)/(sqrt(1+(x^2))) Would I separate these two into 2 ...
- Calculus 1 - Find the derivative of y with respect to x. y=(x^6/6)(lnx)-(x^6/36...
- calculus - sorry to ask a second question so soon, but i'm just not getting this...
- Calculus check and help - Let R be the region bounded by the curves y=lnx^2 and ...
- L'Hopital's rule - Find lim x->1+ of [(1/(x-1))-(1/lnx)]. Here is my work...

More Related Questions