I've tried a lot to solve this and couldn't. it's frustrating. $$ \int_{-1}^1 \frac{xe^{-|x|}}{\sqrt{\cos (x) + 4}} \ dx $$
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5The function looks odd. It’s probably nothing. – Jacky Chong Dec 19 '17 at 16:01
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Use the formula:https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3xdx/439856#439856 – lab bhattacharjee Dec 19 '17 at 16:05
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$$\int_{-1}^1 \frac{xe^{-|x|}}{\sqrt{\cos (x) + 4}} \ dx=0$$ since $x\mapsto \frac{xe^{-|x|}}{\sqrt{\cos (x) + 4}} $ is an odd function – Guy Fsone Dec 19 '17 at 18:38
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Hint
You are integrating $\int_{-1}^1 f(x) dx,\;$ but note that $$ f(-x) = \frac{-xe^{-|-x|}}{\sqrt{\cos (-x) + 4}} = -\frac{xe^{-|x|}}{\sqrt{\cos (x) + 4}} = -f(x), $$ so $f$ is an odd function. Then, substitute $y = -x$ to get $$ \int_{-1}^0 f(x) dx = - \int_1^0 f(-y) dy = - \int_1^0 - f(y) dy = \int_1^0 f(y) dy = -\int_0^1 f(y) dy $$
Can you take it from here?
