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Find the area below the graph of...
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\[y=(1-2cosx)(3+2cosx)e ^{1/2x+cosx}\] 0 (less then and equal to) x (which is less and equal to) 2pi
1/2x+cosx= (1/2)x +cosx ?
should be (1/2)x + cosx
\[\large{\int\limits_{0}^{2 \pi}(1-2cosx)(3+2cosx)e^{(\frac{1}{2}x+cosx)}dx}\]
Yes
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\[\int\limits\limits\limits (1−2cosx)(3+2cosx)e ^{(x/2)+cosx} dx = -4e^{(2cosx+x)/2}sinx-2e^{(2cosx+x)/2}+C\]
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