Question

∫θcosθdΘu=θdv=cosΘdΘdu=dΘv=sinΘ
I got:
∫ΘcosThetadΘ=sin2Θ+cosΘ+C
Is this correct? Thanks:)

Answer 1

\[\int\theta\cos\theta d\theta\]is the problem?

Answer 2

yes. u=theta dv=costheta dtheta

Answer 3

I'm don't think that;'s quite right, but let me do it\[u=\theta\implies du=d\theta\]\[dv=\cos\theta d\theta\implies v=\sin\theta\]\[\int\theta\cos\theta=\theta\sin\theta-\int\sin\theta d\theta=\theta\sin\theta+\cos\theta\]

Answer 4

...+C

Answer 5

do you follow?
do you see your mistake?

Answer 6

But \[\Theta(\sin \Theta)=\sin^2\Theta\]
and the integral of sin=(-cos)??

Answer 7

\[\theta\sin\theta\neq\sin^2\theta\]

Answer 8

and I applied the fact that integral of sin is -cos by changing the - to a + between the two integrals in my final answer

Answer 9

ok, so I just did my algebra wrong by combining the sin theta and theta.

Answer 10

yeah, that's a no-no...

Answer 11

there is no identity I know of for\[x\sin x\]

Answer 12

Got you, Appreciate your help. I'm trying to study for Calc2 in Aug, lol.

Answer 13

Great, good luck. I hope you come to Open Study for help :)