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Evaluate the line integral (x2+y2)dx+2xydy where is the path of the semicircular arc of the circle x2+y2=64 starting at (8,0) and ending at (−8,0) going counterclockwise.

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\[512\int\limits_{0}^{\pi} \cos^2t sint - \sin^3t\] im stuck at how to integrate this.
use\[u=\cos t\]and\[\sin^2t=1-\cos^2t\]

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Other answers:

so du =-sinx
how did you get cos^2t*sint for x^2+y^2dx ?
i keep getting confused but heres where i started \[\int\limits_{0}^{\pi} ((8cost)^2 + (8sint)^2)(-8sint) + 2(8cost)(8sint)(8cost)dt\]
yes, and that can be written\[\int_0^\pi512(\sin^2t+\cos^2t)(-\sin t)dt+\int_0^\pi512(2\cos^2t\sin t) dt\]
\[-512\int\limits_{0}^{?\pi} sint dt +1024\int\limits_{0}^{\pi} \cos^2tsint\] which becomes this
yes, which you can recombine to form this\[512\int_0^\pi2\cos^2t\sin t-\sin tdt\]
so would you use u=cost and du=-sint ?
or could you do \[sint(\cos^2t-1)\] which becomes \[\sin^3t\] ?
and how to you propose to integrate that?
oh yeah.haha.
\[-512\int\limits_{1}^{-1}- 2u^2du - du \] ?
i mean + du
yeah, looks good
oh wait, how did 512 get negative?
im confused because du=-sint and the sint is positive so (-1/-1) would need to be multiplied right?
yeah, but you already covered that by changing thethe signs og the terms
however the double du notation is highly unorthodox and looks screwy to me, so I'd avoid it
\[512\int_0^\pi2\cos^2t\sin t-\sin tdt\]\[u=\cos t\implies du=-\sin tdt\]\[512\int_{1}^{-1}- 2u^2du +\int_1^{-1} -\sin tdt\]
sorry I messed up the bounds on the last integral, they should be 0 to pi
and the answer is 1706.66666666667?
nevermind thats wrong.
\[512\int_0^\pi2\cos^2t\sin t-\sin tdt\]\[u=\cos t\implies du=-\sin tdt\]\[512\int_{1}^{-1}- 2u^2du +\int_0^{\pi} -\sin tdt\]let me see what I get...
\[512\left(\left.-\frac23u^3\right|_1^{-1}+\left.\cos t\right|_0^\pi\right)\]\[512(\frac43-2)=512(-\frac23)\]but I always could have made a mistake...
its right! yay!
sweet, I just hope you find your (likely algebra-based) mistake
thats what i got to too! thanks so much!

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