Yes.

No. The [imath] +C [/imath] is only used for indefinite integrals in order to note that integrals are only unique up to additive terms because [imath] (f(x)+C)'=f'(x). [/imath] As soon as we consider definite integrals like

[math] \int_0^3 x^2\,dx =\left[\dfrac{x^3}{3}\right]_0^3 =\dfrac{3^3}{3}-\dfrac{0^3}{3}=9[/math]we don't need [imath] C [/imath] anymore because we have the exact values at both ends of the integral, other than [math] \int x^2 \,dx = \dfrac{x^3}{3}+C[/math] where we only know the anti-derivative, but not the values at its ends. The [imath] +C [/imath] means literally "plus some **c**onstant - therefore the [imath] C [/imath] - that we cannot know yet".

You can do this, but you don't have to. E.g. if we want to integrate [imath] \int_0^\pi\sin(\pi-x)\,dx [/imath] then we must somehow deal with the [imath] \pi -x [/imath] term. We can do it by using the addition theorem for the sine function, or we can substitute [imath] u=\pi -x [/imath] with [imath] dx=-du [/imath] and [imath] u(0)=\pi\, , \,u(\pi)=0 [/imath] and calculate

[math]\begin{array}{lll}
\int_0^\pi\sin(\pi-x)\,dx&=\int_{x=0}^{x=\pi}\sin(\pi-x)\,dx\\[6pt]
&=\int_{u=\pi}^{u=0}\sin(u)\,(-du)\\[6pt]
&=-\int_{u=\pi}^{u=0}\sin(u)\,du\\[6pt]
&=-\left(-\int_{u=0}^{u=\pi}\sin(u)\,du\right)\\[6pt]
&=\int_{u=0}^{u=\pi}\sin(u)\,du\\[6pt]
&=\left[-\cos(u)\right]_{u=0}^{u=\pi}\\[6pt]
&=-\left(\cos(\pi)-\cos(0)\right)\\[6pt]
&=-(-1-1)=-(-2)=2
\end{array}[/math]

You can do this by the method we are talking about in this thread and that I just used to directly get [imath]\int_0^3 (3 - x)^2 \,dx=\dfrac{27}{3}=9 [/imath] or do it step by step (with a corrected coefficient)

[math]\begin{array}{lll}
\int_0^3 (3 - x)^2 \,dx&= \int_0^3 (9-6x+x^2)\,dx\\[6pt]
&=9\int_0^3\,dx-6\int_0^3 x\,dx+\int_0^3x^2\,dx\\[6pt]
&=9\left[x\right]_0^3 -6\left[\dfrac{x^2}{2}\right]_0^3+\left[\dfrac{x^3}{3}\right]_0^3\\[16pt]
&=9\cdot (3-0) -6 \cdot\left(\dfrac{3^2}{2}-\dfrac{0^2}{2}\right)+\left(\dfrac{27}{3}-\dfrac{0}{3}\right)\\[6pt]
&=9\cdot 3- 6\cdot 4.5 + 9\\[6pt]
&=9
\end{array}[/math]