Calc II, integrals

[Blitze105]

Hello,
I have two questions that i would appreciate some assistance with. I do NOT have the answers to these problems as they are even numbers.

the first is: Find the integral of
(t)sin(t) dt

I have tried the reverse product rule... however my answer seems very strange.

The second is:
pe^-.1p dp

I know i have no work on these problems, and thats not usually how i post questions. However, i am at a loss as how to start the first problem since my idea gave me such an odd answer. As for the second.. does this problem have a trick or is it just using a decimal to throw me off?

 

[soroban]

Hello0, Blitze105!

These require Integration By Parts, which I assume is your "reverse product rule".
I would expect that the odd-numbered problems are the same type.
. . So exactly where is your difficulty?

\(\displaystyle I \;=\;\int t\sin t\,dt\)

. . \(\displaystyle \begin{array}{cccccccc}u &=& t & & dv &=&\sin t\,dt \\ du &=& dt & & v &=& -\cos t \end{array}\)

\(\displaystyle I \;=\;-t\cos t + \int\cos t\,dt\)

\(\displaystyle I \;=\;-t\cos t + \sin t + C\)

\(\displaystyle I \:=\:\int pe^{-0.1p}\, dp\)

. . \(\displaystyle \begin{array}{cccccccc}u &=& p && dv &=& e^{-0.1p}\,dp \\ \\[-3mm] du &=& dp && v &=& -10e^{-0.1p} \end{array}\)

\(\displaystyle I \;=\;-10pe^{-0.1p} + 10\int e^{-0.1p}\,dp\)

\(\displaystyle I \;=\; -10pe^{-0.1p} - 100e^{0.1p} + C\)

\(\displaystyle I \;=\;-10e^{-0.1p}(p + 10) + C\)

 

[Blitze105]

Well, my problem was mainly that my first answer didn't look accurate to my out of practice eyes.

As for the 2nd, I was making a simple and foolish math mistake
Thank you for your help i truly appreciate it!