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.
Best Answer
Generally, integration by parts states that under some nice conditions on differentiable functions $u(x),v(x)$, you have $$ \int u(x) dv(x) = u(x) v(x) - \int v(x) du(x). $$
When your integrals are definite, both sides end up evaluated on the same interval, in other words, $$ \begin{split} \int_a^b u(x) dv(x) &= \left[u(x) v(x)\right]_a^b - \int_a^b v(x) du(x) \\ &= u(b) v(b) - u(a)v(a) - \int_a^b v(x) du(x) \end{split} $$
In your specific case, the added term $$u(b) v(b) - u(a)v(a) = 0$$ which makes things even simpler...