Integrals that simplify by parity

Why the rule holds. The substitution $x\mapsto-x$ gives $\int_{-a}^0f(x)\,dx=\int_0^af(-u)\,du$. For an odd function this is $-\int_0^af$, which cancels the other half. For an even function it equals $\int_0^af$, which doubles it.

1. $x\mapsto x^5e^{x^2}$ is odd (odd power times even function), so the integral is $0$.

2. $x^3$ is odd, so it contributes $0$. $3x^2$ is even: $2\int_0^13x^2\,dx=2$. The result is $2$.

3. $x\cos x$ is odd, so it contributes $0$. For the other term, $\sin^2x=\frac{1-\cos2x}2$ gives $\int_{-\pi}^\pi\sin^2x\,dx=\pi$. The result is $\pi$.

Remark. Recognising symmetry can turn an intimidating integral, like the first one, into zero in one line.