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Mathematics 12 Online
OpenStudy (anonymous):

integrate cos(x/3) from 0 to pi using u substitution

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

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

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?

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

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

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)?

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

OpenStudy (anonymous):

(3(sqrt(3))/2

OpenStudy (anonymous):

\[(3\sqrt(3))/2\]

OpenStudy (amistre64):

yep

OpenStudy (anonymous):

woot

OpenStudy (anonymous):

thanks all

OpenStudy (amistre64):

yw

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