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OpenStudy (anonymous):
Sin(2x) is a trig identity which expands out to 2sinxcosx.
OpenStudy (anonymous):
Okay then should I set u=sin2x? and dv=dx
OpenStudy (anonymous):
No. if sin2x = 2sinxcosx, you can cancel out the sinx
OpenStudy (anonymous):
Okay so I set u=2sinxcosx?
OpenStudy (anonymous):
There is no u substitution.
sin2x is a trig identity
sin2x is EQUAL to 2sinxcosx.
This means that your equation can be rewritten as 2sinxcosx/sinx.
You can cancel out the sinx
So you end up with 2cosx
And then take the integral of 2cosxdx
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OpenStudy (anonymous):
Okay....Then u should be cosxdx?
OpenStudy (anonymous):
Where is u coming from?
OpenStudy (anonymous):
From the last part integral of 2cosxdx
OpenStudy (anonymous):
When I take the integral of that should u=cosx?
OpenStudy (anonymous):
Where is u?
The integral is just
\[\int\limits_{}^{}2cosxdx\]
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OpenStudy (anonymous):
I need to find the anti-derivative to the integral of 2cosxdx
OpenStudy (anonymous):
Yes.
OpenStudy (anonymous):
So shouldnt I set u=cosx
OpenStudy (anonymous):
\[1/2\int\limits_,2cosxdx\] u=cosx, du=sinxdx
OpenStudy (anonymous):
You don't need u at all.
\[\int\limits_{}^{}2cosxdx = 2\int\limits_{}^{}cosxdx\]
Which equals 2sinx + c.