# Calculus

posted by Anonymous

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

integral xcosxdx
x cosx
1 sinx
0 -cosx

integral e^2x(sinx)dx
u=e^2x dv=sinxdx
du=2e^2x v=-cosx
-e^x(cosx)+integral(2e^2x)(cosx)

integral (x^2)(e^2x)
x^2 e^2x
2x (1/2)e^2x
2 (1/4)e^2x
0 (1/8)e^2x

1. drwls

integral x^3(lnx)dx
u=lnx dv=x^3dx
du=1/x v=x^4/4

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.

2. Anonymous

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

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