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There is no problem. See the comment by @BabyDragon above.

Note that the integrand satisfies $f(\frac{\pi}{2}-x) = -f(\frac{\pi}{2}+x)$ over the range of integration, so the answer is zero.

There is no problem. See the comment by @BabyDragon above.

Note that the integrand satisfies $f(\frac{\pi}{2}-x) = -f(\frac{\pi}{2}+x)$ over the range of integration, so the answer is zero.

Regarding the last remark, note that for sufficiently smooth $f,u$, we have $$ \int_{u(a)}^{u(b)} f(x) dx = \int_a^b f(u(t))u'(t) dt$$

With $a=0$, $b=1$, $f(x) = x$, and $u(t) = t^2-t$, this will result in

$$\int_0^0x dx = \int_0^1 (t^2-t)(2t-1)dt = 0$$

In particular, note how the change of variables affects the integration bounds.

There is no problem. See the comment by $BabyDragon above.

Note that the integrand satisfies $f(\frac{\pi}{2}-x) = -f(\frac{\pi}{2}+x)$ over the range of integration, so the answer is zero.

There is no problem. See the comment by @BabyDragon above.

Note that the integrand satisfies $f(\frac{\pi}{2}-x) = -f(\frac{\pi}{2}+x)$ over the range of integration, so the answer is zero.

The problem is that $\sin$ is not injective over the range of integration. You need to split the integral into $\int_0^{\frac{\pi}{2}}$ and $\int_{\frac{\pi}{2}}^\pi$.

Note that the integrand satisfies $f(\frac{\pi}{2}-x) = -f(\frac{\pi}{2}+x)$ over the range of integration, so the answer is zero. However, you do need to exercise care when the substitution is not injective.

There is no problem. See the comment by $BabyDragon above.

Note that the integrand satisfies $f(\frac{\pi}{2}-x) = -f(\frac{\pi}{2}+x)$ over the range of integration, so the answer is zero.