integrate cos(x/3) from 0 to pi using u substitution
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OpenStudy (anonymous):
u = x/3 ???
OpenStudy (amistre64):
yep
OpenStudy (anonymous):
du = (1/6)x^2
OpenStudy (amistre64):
derivatives dont work like that ....
OpenStudy (amistre64):
du = 1/3 dx
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OpenStudy (amistre64):
3du = dx
OpenStudy (anonymous):
oh, I was trying to integrate
OpenStudy (amistre64):
0 < x < pi
0/3 < x/3 < pi/3
0 < u < pi/3
OpenStudy (anonymous):
@amistre64 is smart
OpenStudy (amistre64):
integrate [0,pi] cos(x/3) dx
replace all the parts
integrate [0,pi/3] cos(u) 3 du
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OpenStudy (amistre64):
3sin(pi/3) - 3 sin(0)
OpenStudy (amistre64):
im assuming you know how to integrate cos(u) of course
OpenStudy (anonymous):
getting there
OpenStudy (amistre64):
strategy; replace all x parts by u parts
OpenStudy (anonymous):
how is that step written, without the upper and lower bounds?
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OpenStudy (amistre64):
we have the upper and lower bounds, so im not sur what you mean
OpenStudy (anonymous):
after we choose the substitution, find its derivative, then the next step looks like what...to me, it doesn't make sense to put the upper and lower bounds on it...am I wrong?
OpenStudy (amistre64):
the bounds on x are:
0 < x < pi
right?
but u = x/3, or 3u = x
0 < 3u < pi ; solve for u, divide off the 3
0 < u < pi/3
OpenStudy (amistre64):
we now have the bounds in terms of u
OpenStudy (anonymous):
ok...processing
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OpenStudy (anonymous):
\[3\int\limits_{0}^{\pi/3}\cos(u)du\]
OpenStudy (amistre64):
yeah
OpenStudy (anonymous):
ok, now I integrate that using indefinite...then subtract upper bounds from lower bounds
OpenStudy (amistre64):
yeah, integration hasnt changed ... the key here is in converting all the x parts to u parts.
OpenStudy (anonymous):
gotcha
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OpenStudy (anonymous):
how is this next step written
OpenStudy (amistre64):
its integration, youve already covered that before you delve into subsitutions ...
OpenStudy (amistre64):
youve even mentioned how to work it ....
OpenStudy (amistre64):
and its a definite integral, its defined by the boundarys.
indefinite has no bounds
OpenStudy (anonymous):
so, do I just write 3sin(u)?
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OpenStudy (amistre64):
almost, plug in your bounds ....
OpenStudy (amistre64):
you get a value, not a function ....
OpenStudy (anonymous):
right, I shouldn't write that last part by itself
OpenStudy (amistre64):
3sin(u) [evaluated at: 0 and pi/3]
and ive already posted above what it looks like ....
OpenStudy (anonymous):
ok, processing
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OpenStudy (anonymous):
(3(sqrt(3))/2
OpenStudy (anonymous):
\[(3\sqrt(3))/2\]
OpenStudy (amistre64):
yep
OpenStudy (anonymous):
woot
OpenStudy (anonymous):
thanks all
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