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Mathematics 16 Online
OpenStudy (anonymous):

integral question

OpenStudy (anonymous):

\[\int\limits_{}^{} 2r* \sin(2r)\]

Parth (parthkohli):

what do you think about this one?

OpenStudy (anonymous):

We have to use integration by parts

Parth (parthkohli):

that's correct

OpenStudy (anonymous):

\[\int\limits_{}^{}usin(u)\]

OpenStudy (anonymous):

u being 2r

Parth (parthkohli):

no need for unnecessary substitution.

OpenStudy (anonymous):

??

OpenStudy (anonymous):

oh ok

Parth (parthkohli):

no need to substitute it. just take the 2 out.

OpenStudy (anonymous):

is the antiderivative of sin(2r) is -1/2(cos(2r))?

Parth (parthkohli):

yes!

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

r/2 * cos(2r) + cos(2r)/2

OpenStudy (anonymous):

something went wrong >.<

Parth (parthkohli):

\[\int 2r\sin(2r)dr = -r\cos(2r) -\int -\cos(2r)dr\]

OpenStudy (anonymous):

wait let's break this apart

OpenStudy (anonymous):

uv= r * (-1/2) cos(2r)?

Parth (parthkohli):

There's a 2 too.

OpenStudy (anonymous):

how?

OpenStudy (anonymous):

where?

Parth (parthkohli):

2r sin(2r)

OpenStudy (anonymous):

OH

Parth (parthkohli):

oops

Parth (parthkohli):

u = 2r dv = sin(2r) dr

OpenStudy (anonymous):

Thank you

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