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OpenStudy (anonymous):
check my work, integrals
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OpenStudy (anonymous):
\[\int\limits_{0}^{\pi/8}\sin2xdx\]
OpenStudy (anonymous):
=-cos2x -cos(2(pi/8-0)) is that right
OpenStudy (ipwnbunnies):
No, that's incorrect. o.o
OpenStudy (anonymous):
for some reason not getting right answer
OpenStudy (luigi0210):
There's something missing..
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OpenStudy (anonymous):
ok wha
OpenStudy (anonymous):
Try u = 2x before doing the integral.
OpenStudy (anonymous):
@Vandreigan havent gotten that far yet
OpenStudy (luigi0210):
Well the general should be \(\LARGE -\frac{1}{2} cos~2x+C\)
OpenStudy (anonymous):
and you drop the c, correct?
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OpenStudy (luigi0210):
Yea, drop it, plug in your limits and use the fundamental theorem
OpenStudy (anonymous):
so -1/2 cos(2(pi/8-0))
OpenStudy (luigi0210):
Do you know the theorem?
OpenStudy (anonymous):
u = 2x , du= 2dx
OpenStudy (anonymous):
\[\int\limits_0^{\pi/8} \sin(2x)dx = \frac{1}{2} \int\limits_0^{\pi/4} \sin(u)du\] if u = 2x, du = 2dx Now try the integral.
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OpenStudy (anonymous):
fb-fa
OpenStudy (anonymous):
2+2=4
OpenStudy (anonymous):
\[-\sqrt{2}/4\]
OpenStudy (anonymous):
that not right either looking for
OpenStudy (anonymous):
\[(2-\sqrt{2})/4\]
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OpenStudy (anonymous):
?
OpenStudy (anonymous):
Cos(0) = 1, not 0
OpenStudy (loser66):
|dw:1397941181073:dw|
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