Reduction formula for definite integral of $\sin^2(x)$.

Question

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.

Answer 1

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, $$ \begin{split} \int_a^b u(x) dv(x) &= \left[u(x) v(x)\right]_a^b - \int_a^b v(x) du(x) \\ &= u(b) v(b) - u(a)v(a) - \int_a^b v(x) du(x) \end{split} $$

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In your specific case, the added term $$u(b) v(b) - u(a)v(a) = 0$$ which makes things even simpler...

Answer 2

Note that you have a definite integral, with limits $0$ and $\pi$ the term that depends on $x$ only cancels at both those values

Answer 3

For $n>1,$

$$\sin^{n-1}x\cos x\big|_0^\pi=0$$

For odd function likely $f(x)=\sin^3x$

we can prove $$\int_{-a}^af(x)\ dx=0$$ using Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$.