I've been doing some homework with Dirac delta, and while evaluating a given integral, I noticed an 'error' in my approach that I can't quite figure out.
$$\int_{-\pi/2}^{\pi/2} \cos^2\theta ~ \delta(\sin\theta) ~d\theta.$$
I figured a potentially easy solution would involve identity $$cos^2\theta=1-sin^2\theta$$ Thus, expanding the integral
$$\int_{-\pi/2}^{\pi/2} (1-sin^2\theta) \cdot \delta(sin\theta) d\theta$$
$$\int_{-\pi/2}^{\pi/2}\delta(sin\theta) d\theta -\int_{-\pi/2}^{\pi/2} sin^2\theta \cdot\delta(sin\theta) d\theta$$
$$\int_{-\pi/2}^{\pi/2}\delta(sin\theta) d\theta -\int_{-\pi/2}^{\pi/2} sin\theta \cdot sin\theta \cdot\delta(sin\theta) d\theta$$
Now I assumed since,
$$x \delta(x) = 0$$
That
$$sin\theta\delta(sin\theta) = 0$$
as well, which would lead to
$$\int_{-\pi/2}^{\pi/2}\delta(sin\theta) d\theta - 0$$ $$\int_{-\pi/2}^{\pi/2}\delta(sin\theta)d\theta = 1$$
However, I believe the initial integral is supposed to evaluate to zero (at least according to https://www.wolframalpha.com/input?i2d=true&i=Integrate%5BPower%5Bcos%2C2%5D%5C%2840%29x%5C%2841%29%CE%B4%5C%2840%29sinx%5C%2841%29%2C%7Bx%2C-Divide%5B%CF%80%2C2%5D%2CDivide%5B%CF%80%2C2%5D%7D%5D)
So I deduced
$$\int_{-\pi/2}^{\pi/2} cos^2\theta \cdot \delta(sin\theta) d\theta \neq \int_{-\pi/2}^{\pi/2} (1-sin^2\theta) \cdot \delta(sin\theta) d\theta$$
However, I can't quite understand why.
I'm relatively new to dirac-delta integrations, so if anyone would be able to explain as to why we can't utilize the identity and expand, it would be much appreciated!
Hint: A simpler method is integration by substitution:
$$\int_{-\pi/2}^{\pi/2}\! \mathrm{d}\theta~ \cos^2\theta~\delta(\sin\theta)~\stackrel{s=\sin\theta}{=}~\int_{-1}^1\! \mathrm{d}s~ \sqrt{1-s^2}~\delta(s)~=~1. $$