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I used substitution and got that: $$\int_0^\pi \sin x \cdot P_n(\cos x ) \, dx=0$$ where $P_n$ is the $n$-th Legendre polynomial.

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  • $\begingroup$ What integral you were calculating? $\endgroup$ – Mhenni Benghorbal Jul 23 '13 at 11:02
  • $\begingroup$ Are you calculating $\int_{-1}^{1}P_0(t)P_n(t)dt $? $\endgroup$ – Mhenni Benghorbal Jul 23 '13 at 11:13
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Recall that Legendre polynomials are defined as orthogonal polynomials on $[-1,1]$ with weight function $w(x)=1$. In other words, we have by definition $$(P_m,P_n)=\int_{-1}^1P_m(x)P_n(x)dx\sim \delta_{mn}.$$ But, since $P_0(x)=1$, your integral is equal to $(P_0,P_n)$ and therefore it vanishes whenever $n>0$. For $n=0$, however, it is equal to $2$.

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  • $\begingroup$ Thank you, but why is it equal to 2, I mean if my integral is $(P_0,P_n)$ then it should be equal to 1, as $(P_0,P_0)=1$ is the only one that survives? $\endgroup$ – user66906 Jul 23 '13 at 11:02
  • $\begingroup$ @Lipschitz Legendre polynomials are orthogonal, not orthonormal. $\endgroup$ – Start wearing purple Jul 23 '13 at 11:03
  • $\begingroup$ you are right, sorry! thank you very much $\endgroup$ – user66906 Jul 23 '13 at 11:04
  • $\begingroup$ @Lipschitz You are welcome! In fact, in front of the Kronecker symbol there is a factor $2/(2n+1)$. But here you don't even need to know it, just integrate $1$ from $-1$ to $1$. Of course, the result will reproduce $2/(2\cdot 0+1)$. $\endgroup$ – Start wearing purple Jul 23 '13 at 11:07
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If you look at the Power reduction formulas for the cosine, you see that all monomials in your Legendre polynomial involve only cosines of integer multiples of $x$. These are odd functions on the interval $[0,\pi]$ while the sine function is even, thus their product must be odd and the integral is zero.

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  • $\begingroup$ I think the oddness is not quite true: e.g. $\cos2x=1-2\sin^2x$. $\endgroup$ – Start wearing purple Jul 23 '13 at 11:16

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