Compute $$\int_{-\pi/2}^{\pi/2} \frac{x^2\cos(x)}{1+e^x}$$ The first step itself given in the solution is changing $e^x$ to $1/e^x$ . Now as first step is making no sense to me so I didn't post the whole solution. Hope you guys help and tell me if this is some rule for definite integration.
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So you want to evaluate $$\int \limits_\frac{\pi}{2}^{\frac{\pi}{2}}\frac{x^{2}e^{-x}\cos x dx}{1+e^{-x}}$$ – May 26 '16 at 15:15
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Yes how to go about it – Archis Welankar May 26 '16 at 15:18
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The solution substitute $x' = -x$. You can plot the functions in Mathematica to see what's happening (I notice the symmetry in y-axis). Consider both $x^2$ and $e^x$ are even functions. – user202729 May 26 '16 at 15:20
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Maybe they made the substitution $t=-x$? (With the plan of adding that integral to the original integral; that's a trick which I'm sure has been explained in other questions here. Now it's only a matter of finding them...) – Hans Lundmark May 26 '16 at 15:21
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Can anyone show how to evaluate this integral? – Mathxx May 26 '16 at 15:50
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After seeing the comments I now know what's going on Thanks – Archis Welankar May 26 '16 at 16:06
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i have found this approximativly result $$0.4674011002723396547086227499690377838284248518101976566033373440550112056048013107504433509296380580$$ – Dr. Sonnhard Graubner May 26 '16 at 16:38
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No all answers are in $\pi$ so no problem with that. – Archis Welankar May 26 '16 at 16:40
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Here are a few relevant links: http://math.stackexchange.com/questions/609639/an-intriguing-integral-i-int-limits-04-fracdx42x, http://math.stackexchange.com/questions/70974/lesser-known-integration-tricks/691064#691064 – Hans Lundmark May 26 '16 at 17:29
1 Answers
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Let $x=-u$
$$I=\int_{\pi/2}^{-\pi/2}- \frac{u^2 \cos (u)}{1+e^{-u}} du= \int_{-\pi/2}^{\pi/2} \frac{e^u u^2 \cos (u)}{1+e^{u}} du $$
Add the last integral to the original one gives
$$2I =\int_{-\pi/2}^{\pi/2} x^2 \cos x dx$$
Which can be dealt with using integration by part.
lEm
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