How do I integrate (1-t^2)*e^(-[t^2]/2)?

Question

I am completely stuck on this integral.  I know what the answer is, but i don't know the steps i am supposed to take to reach it.

I tried thinking of a clever u substitution and considered even integrating by parts, but every time i think it over it doesn't seem like it will come out well.  Can someone explain it to me?

? · Accepted Answer

If I understand this correctly from the numbers and symbols given:

∫ [(1 - t²)*e^(-t²/2)] dt

= ∫ e^(-t²/2) - t^2[e^-(t²/2)] dt

Let u = -t²/2

Then du = -t dt

And then dt = -du/t

Replace that value found for dt in terms of u:

∫ (t²)(e^u) *(-du/t)

Now cancel out terms, watch - signs and simply to:

∫ te^u du

Go back and observe when we let u = -t²/2
Since √(-2u) and -√(-2u) are like terms, add -√(-2u) to √(-2u) and this becomes 0

∫ (0)e^u du

∫ 0 du = 0 + C

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How do I integrate (1-t^2)*e^(-[t^2]/2)?

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