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