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I'm following a course on analysis and I am supposed to compute the following integral:

$\int_{-1}^{1}x^2\sin^{2019}(x)e^{-x^4}dx$.

I've been trying to use the integration by part but I've been stuck for a while. Any help would be highly appreciated. Thanks

Bernard
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jffi
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  • Use https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 and the given function is odd – lab bhattacharjee Dec 31 '18 at 14:18

2 Answers2

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Just substitute $x=-t$ and you will get that: $$I=\int_{-1}^{1}x^2\sin^{2019}(x)e^{-x^4}dx=-\int_{-1}^{1}t^2\sin^{2019}(t)e^{-t^4}dt=-I$$ $$I=-I\Rightarrow I=0$$

Zacky
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    Oh... right! I can't believe I was actually trying to compute the primitive. Thanks – jffi Dec 31 '18 at 14:26
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    I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up! – Zacky Dec 31 '18 at 14:32
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The integrand is an odd function.

For any odd (integrable) function $f$, we have

$$\int_{-a}^a f = 0$$

Therefore, the integral is equal to $0$.