How do I evaluate $\int_{-\infty}^{\infty}(1-x^{2})e^{-x^{2}}\mathrm{d}x$ using simple calculus? How do I evaluate this integral using simple calculus techniques such as substitution or parts?
$$\int\limits_{-\infty}^{\infty}(1-x^{2})e^{-x^{2}}\mathrm{d}x$$
I would not like any solution involving more advanced techniques such as the gamma function, the error function or polar coordinates.
 A: Integrate by parts $$\int_{-\infty}^{\infty} e^{-x^2} \; dx =
\left. xe^{-x^2}\right|_{-\infty}^{\infty}-\int_{-\infty}^{\infty} -2x^2 e^{-x^2} \; dx.$$  Using the classic "switch to polar" trick for the first integral gives
$$\sqrt{\pi} = 2\int_{-\infty}^{\infty} x^2 e^{-x^2} \; dx.$$  So your integral equals $\sqrt{\pi} - \sqrt{\pi}/2 = \sqrt{\pi}/2.$
A: 
Here, we use "Feynman's Trick" for Differentiating Under the Integral as a  straightforward way to evaluate integrals of the form $\int_{-\infty}^\infty x^{2n}e^{-x^2}\,dx$.  To that end, we proceed.

Let $I(a)$ be the integral defined by 
$$I(a)=\int_{-\infty}^\infty e^{-ax^2}\,dx \tag 1$$
The integral in $(1)$ is a Gaussian integral, which can be evaluated a number of ways (SEE HERE for a way that avoids using polar coordinates or the Gammma function ) to obtain
$$I(a)=\sqrt{\frac{\pi}a} \tag 2$$
Now, differentiating the right-hand side of $(1)$ reveals
$$I'(a)=-\int_{-\infty}^{\infty}x^2e^{-ax^2}\,dx \tag 3$$
while differentiating the right-hand side of $(2)$ reveals
$$I'(a)=-\frac12 \sqrt{\pi}a^{-3/2} \tag 4$$
Upon equating $(3)$ and $(4)$  and setting $a=1$, we find

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty x^2e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}}$$ 

We can continue to differentiate $(1)$ to obtain
$$\begin{align}
I^{(n)}(a)&=(-1)^n \int_{-\infty}^\infty x^{2n}e^{-ax^2}\,dx\\\\
&=(-1)^n\frac{(2n-1)!!}{2^n}a^{-(2n+1)/2}
\end{align}$$
from which we arrive at 
$$\int_{-\infty}^\infty x^{2n}e^{-ax^2}\,dx=\frac{\sqrt{\pi}\,(2n-1)!!}{2^n}\,a^{-(2n+1)/2} \tag 5$$
Setting $a=1$ in $(5)$ yields

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty x^{2n}e^{-x^2}\,dx=\frac{\sqrt{\pi}\,(2n-1)!!}{2^n} }$$

A: $\int_{-\infty}^{\infty} e^{-x^2} dx - \int_{-\infty}^{\infty} x^2e^{-x^2} dx$
Lets tackel this in pieces
$\int_{-\infty}^{\infty} x^2e^{-x^2} dx$ integrate by parts
$u =  dx, dv = xe^{-x^2} dx\\
du = dx, v = -\frac 12 e^{-x^2}$
$-\frac x2 e^{-x^2} |_{-\infty}^{\infty} + \frac 12 \int e^{-x^2} dx \\
-\frac x2 e^{-x^2} |_{-\infty}^{\infty} = 0$
And what is left pairs with the piece we had not yet addressed.
Since the fucntion is even,
$\frac 12 \int_{-\infty}^{\infty} e^{-x^2} dx = \int_{0}^{\infty} e^{-x^2} dx$
You said you don't want polar coordinates, which is actually the easy way to solve this one.
Consider:
$F(t) = \int_0^{\infty} \frac {e^{-t^2(x^2+1)}}{1+x^2} dx\\
F'(t) = \int_0^{\infty} 2t e^{-t^2(x^2+1)} dx$
Note that we are differentiating with respect to t and now with respect to x.
$F'(t) = -2e^-t^2 \int_0^{\infty}  te^{-(tx)^2} dx\\
u = tx\\
du = t dx\\
F'(t) = -2e^-t^2 \int_0^{\infty}  e^{-u^2} du$
So...
$F'(t) = e^-t^2 (-2I)$ where I is a constant (and I is also the number we are looking for.)
The fundamntal theorm of calucus says:
$\int_a^b F'(t) dt = F(b) - F(a)\\
F(b)- F(a) = -2I \int_b^a e^-t^2 dt $
Set $a = 0,$ and $b$ approaches infinity.
$F(\infty) - F(0) = -2I\int_0^infty e^-t^2 dt  = -2 I^2$
$F(0) = \int_0^{\infty} \frac {1}{1+x^2} dx$
Which integrates to an inverse tangent function and equals $\frac \pi 2$
$F(\infty) = \int_0^{\infty} 0 dx = 0$
$F(\infty) - F(0) = -\frac \pi 2 = -2 I^2\\
I^2 = \frac {\pi}4\\
I = \frac {\sqrt{\pi}}2$
A: $$\int_{-\infty}^{\infty}\exp(-x^2)dx = \sqrt{\pi}$$
Substituting $x = \sqrt{a} t$ in here gives:
$$\int_{-\infty}^{\infty}\exp(-a t^2)dx = \sqrt{\frac{\pi}{a}}$$
Put $a = 1+u$ and expand both sides in powers of $u$. The integrand can be expanded as:
$$\exp\left[-(1+u)t^2\right]= \exp(-ut^2)\exp(-t^2)=\left[1-ut^2 +\mathcal{O}(u^2)\right]\exp(-t^2)$$
We also have:
$$\sqrt{\frac{\pi}{1+u}}=\sqrt{\pi}\left(1-\frac{1}{2}u+\mathcal{O}(u^2)\right)$$
Equating the integral to the result up to order $u$ yields:
$$\int_{-\infty}^{\infty}\left(1-ut^2\right)\exp(-t^2)dt =\sqrt{\pi}\left(1-\frac{1}{2}u\right)$$
The desired integral is thus $\frac{\sqrt{\pi}}{2}$.
