When trying to proof the reduction formula: $$I\_n =\frac{n-1}{n}\cdot I\_{n-2}$$ for the definite integral $I\_n:=\int\_{0}^{\pi} \sin(x)^{n} dx$, I tried going about it by reducing the indefinite integral $I\_n:=\int \sin(x)^{n} dx$ to: $$I\_n= \frac{-\sin(x)^{n-1}\cdot \cos(x)}{n}+\frac{n-1}{n}\int \sin(x)^{n-2} dx$$ which was still fine. But, since the new form now has a part which depends on $x$, namely: $$\frac{-\sin(x)^{n-1}\cdot \cos(x)}{n}$$ How do I take the definite integral? And how would I, for example take an integral such as $\int\_{-2}^{2} \sin^3(x)dx$ using the reduced form?

Thanks for an answer and excuse possible inaccuracies in my wording since I'm very new to talking about math in English.