Evaluating $\int_{-\infty}^\infty \sin(x)\cos(x)\ dx$

Question

Why is $$\int_{-\infty}^\infty \sin(x)\cos(x)\thinspace \mathrm{d}x=0?$$ My textbook says it is so by symmetry, but it seems like using that logic, the integral of $\sin(x)$ from $-\infty$ to $\infty$ should also be $0$, but it is not.

Answer 1

$$\int_{-\infty}^\infty \sin(x)\cos(x)\thinspace \mathrm{d}x=$$

$$\int_{-\infty}^0 \sin(x)\cos(x)\thinspace \mathrm{d}x+$$

$$\int_0^{\infty} \sin(x)\cos(x)\thinspace \mathrm{d}x$$

None of the last tow integrals exist, so $$\int_{-\infty}^\infty \sin(x)\cos(x)\thinspace \mathrm{d}x$$

Does not exist.

Answer 2

The only thing we could write is $$\int \sin(x)\cos(x)\,dx=-\frac{1}{2} \cos ^2(x)+C$$ making $$\int_{-a}^{+a} \sin(x)\cos(x)\,dx=0$$