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)
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