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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.

gt6989b
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psyph
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  • You are evaluating definite integral so limits should be applied everywhere during the integration. The term that you are mentioning is equal to zero for both $x=0$ and $x=\pi$ and therefore vanishes leaving you with the recurrence formula that you are trying to prove. – Saša May 14 '19 at 14:41

3 Answers3

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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...

gt6989b
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Note that you have a definite integral, with limits $0$ and $\pi$ the term that depends on $x$ only cancels at both those values

Andrei
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  • I did notice that but I didn't unterstand in which way I would put my limits into the part of the equation thats not the integral. Thats why asked for an explanation using a different example, which, unfortunately I chose poorly, not thinking about that it automatically equals 0. What I'm trying to understand is, how would I use the reduction formula in a case such as $\int_{-1}^{2} sin^3{x}$, where I have to get a value for a definite integral using two limits, but having a part that depends on one value for x. – psyph May 14 '19 at 14:53
  • As mentioned in other answers, $\sin^{n-1}x\cos x\big|0^\pi=0$. If your limits are not $0$ and $\pi$, the formula that you have for induction is not valid. So in your case, with limits $2$ and $-1$, the formula would be $$I_n=\frac{n-1}n I{n-2}-\frac{\sin^{n-1}x\cos x}{n}\big|_{-1}^2$$The last term evaluates to a function that depends only on $n$. – Andrei May 14 '19 at 15:00
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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 $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$.

lab bhattacharjee
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