Why did this integral return $0 = 0$ when applying $2$ Pythagorean identities instead of $1$? Had this question on a homework the other day $\int \sin^3x \cos^2x \ \mathrm dx$ pretty simple right but upon my first try through it I turned $\int \sin x \sin^2x \cos^2x \ \mathrm dx$ into $\int \sin x(1-\cos^2x)(1-\sin^2x)\ \mathrm dx$
Now when I broke this all down I got 0 = 0 which I thought was really interesting because this all seemed like I wasnt really breaking any rules?
But if I just do $\int \sin x(1-\cos^2x)\cos^2x \ \mathrm dx$ and substitute then you get an answer really easily. I guess im just curious as to if something above was not allowed? I checked the algebra after breaking it down a bunch and it all seemed sound. Sorry just a little curious as to what it even means
Edit:
If you break down the 2 Pythagorean identities equation you eventually get
$\int \sin^3x \cos^2x \ \mathrm dx = \int \sin x \ \mathrm dx - \int \sin^3x \ \mathrm dx- \int \sin x \cos^2x \ \mathrm dx + \int \sin^3x \cos^2x \ \mathrm dx$. If you subtract the $\int \sin^3x\cos^2x\ \mathrm dx$ from both side you get
$$0 = \int \sin x \ \mathrm dx  - \int \sin^3x \ \mathrm dx - \int \sin x\cos^2x  \ \mathrm dx$$
which then breaks down to $0 = 0$
Just curious as to how that's possible or if I messed up a rule?

You didnt do anything wrong. If you start with an expression and then using algebraic manipulation you get $0=0$ (or $n=n$) that means the expression you began with is true or this case it means you didnt make any mistakes
Notice that the value of integral is not zero
