Question

Am I missing a negative somewhere in one of the steps to solving this problem: http://imgur.com/ou8a8

Answer 1

I apologize for my atrocious hand writing

Answer 2

The answer says I am that last integral on the left on the bottom is supposed to be negative and I am supposed to add it to the other side and then divide by 2 to get the final answer

Answer 3

*on the right

Answer 4

those cyclic problem

Answer 5

?

Answer 6

Let me know if I need to clarify my methodology anywhere

Answer 7

I am going write your work down, the blur is giving me headache

Answer 8

ok

Answer 9

The problem is example 8 here: http://tutorial.math.lamar.edu/Classes/CalcII/IntegrationByParts.aspx

Answer 10

I chose the opposite for u and dv though

Answer 11

Answer should still be the same

Answer 12

At least according to paul u and dv are interchangable on this problem.

Answer 13

I found your error
\[e^{\theta }\text{Sin}[\theta ]-\left(\int e^{\theta }\text{Sin}[\theta ]\right)\]
\[\int e^{\theta }\text{Cos}[\theta ]=e^{\theta }\text{Sin}[\theta ]-\left(-e^{\theta }\text{Cos}[\theta ]+\int e^{\theta }\text{Cos}[\theta ]\right)\]
\[\int e^{\theta }\text{Cos}[\theta ] =e^{\theta }\text{Sin}[\theta ]+e^{\theta }\text{Cos}[\theta ]-\int e^{\theta }\text{Cos}[\theta ]\]

Answer 14

You weren't factoring - before the integral

Answer 15

which integral, last one?

Answer 16

\[e^{\theta }\text{Sin}[\theta ]-\left(\int e^{\theta }\text{Sin}[\theta ]\right)\]

Answer 17

hmm

Answer 18

\[(−eθCos[θ]+∫eθCos[θ])\]

Answer 19

Where why is there no negative there before the distribution

Answer 20

if V is -cosθ

Answer 21

hmm

Answer 22

I am with you up to here: \[eθSin[θ]−(∫eθSin[θ])\]

Answer 23

then I make u = eθ and dv = Sin[θ]

Answer 24

\[\left(-e^{\theta }\text{Cos}[\theta ]-\int -e^{\theta }\text{Cos}[\theta ]\right)\]

Answer 25

dv= sin theta
v= - cos[theta]

Answer 26

Ahhh

Answer 27

Ok, I see it now

Answer 28

Anymore than 2 negatives gets tricky lol

Answer 29

Thanks!

Answer 30

Those are called cyclic, gave me a lot of trouble while I was taking calc II