Integration by parts

1. Jun 3, 2006

suspenc3

hi..im new to this topic..can someone check to see if this is right?

$$\int (xe^-^x)dx = \int udV = uV - \int Vdu$$
$$=x(-e^-^x)- \int -e^-^x$$
$$=-xe^-^x-e^-^x+C$$

thanks

2. Jun 3, 2006

arildno

Why not check it yourself?

Last edited: Jun 3, 2006
3. Jun 3, 2006

suspenc3

wow..why didnt i think of that haha

THanks

4. Jun 3, 2006

arildno

Yeah, it's right

5. Jun 3, 2006

suspenc3

Ok, I have another one im trying to figure out...but i keep getting 0!

$$\int_{0}^{\pi} tsin3t dt = udV \right]_{0}^{\pi} - \int_{0}^{\pi}vdu$$

work...

$$=t(sin3t) \\right]_{0}^{\pi} - \int_{0}^{\pi}vdu$$

$$=\pi (sin3\pi) - \int_{0}^{\pi} \frac{-1}{3} cos3t(1)dt$$

$$=0 - \frac{1}{sin3t} \int_{0}^{\pi}z dz$$..
let z = cos3t
and then use substitution to get $$\frac{dz}{sin3t} = -3dt$$

$$0-\frac{1}{sin3t} \frac{z^2}{2}$$

$$=\frac{-(cos3t)^2}{2sin3t}$$
this must be wrong..i solved it out a bit more..but keep getting zero, where did i go wrong?

Last edited: Jun 3, 2006
6. Jun 3, 2006

arildno

Well, what you have written is just nonsense.
As a novice, it is safer for you to do these problems like this:
$$u(t)=t\to\frac{du}{dt}=1,\frac{dv}{dt}=\sin(3t)\to{v}(t)=-\frac{1}{3}\cos(3t)$$

Use these relations in the integration by parts formula.

7. Jun 3, 2006

suspenc3

hrmm..i already knew these relations..i still dont see what i did wrong..i followed it just how they do it in the book, my LaTeX skills arent very good, is it just hard to follow?

8. Jun 3, 2006

arildno

well, you didn't compute the uv-term correctly.

9. Jun 3, 2006

suspenc3

o wait..it is :$$uv - \int_{0}^{\pi}vdu$$
i have $$udv - \int_{0}^{\pi}vdu$$ in my work..my bad

10. Jun 3, 2006

suspenc3

giving me $$\frac{\pi}{3}$$ which i think is right..thanks for the help

11. Jun 4, 2006

HallsofIvy

I would strongly recommend that you write out exactly what "u", "v", "du", and "dv" are!