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I am struggling with this prove integral is $0$ question, I would really appreciate anyone that could help.

The questions asks to prove that for m ≠ n: $$\int_{-1}^1\frac{\cos nx\cos mx}{\sqrt{1 − x^2}}\ dx = 0$$

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I know the integral is twice the integral from $0$ to $1$, but I have no idea how to prove it.because cos is not an odd function so the integral over a symmetric interval is not $0$...

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  • $\begingroup$ Something is wrong here. If $n=0$ and $m=1$, the integrand $\cos x/\sqrt{1-x^2}$ is strictly positive on the interval $-1\lt x\lt 1$, so the integral cannot be $0$. $\endgroup$ – Barry Cipra Feb 2 '17 at 11:14
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A start:

$\cos a \cos b =(\cos(a-b)+\cos(a+b))/2$.

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  • $\begingroup$ Please see my comment below the OP. $\endgroup$ – Barry Cipra Feb 2 '17 at 11:21
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Hint: $\cos{nx}\cdot \cos{mx} = \dfrac{1}{2}(\cos{(n-m)x} + \cos{(n+m)x})$.

You are on the right path. Now using this identity you can split this into a sum of two integrals, and work the rest using substitution on each integral

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  • $\begingroup$ Please see my comment below the OP. $\endgroup$ – Barry Cipra Feb 2 '17 at 11:22
  • $\begingroup$ Yes, I see, it was corrected beforehand that probably the OP needs to change $\cos{nx}$ by $\cos{\pi nx}$, etc... $\endgroup$ – Arnold Frenzy Feb 3 '17 at 1:00
  • $\begingroup$ I thought so too at first. (I made, then deleted, a comment to that effect.) But $\int_{-1}^1{\cos\pi x\over\sqrt{1-x^2}}dx$ is easily seen to be negative. $\endgroup$ – Barry Cipra Feb 3 '17 at 11:26
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It's no wonder you are struggling: What you are trying to prove is simply not true in general.

The quickest way to see this is to consider the special case $n=0$, $m=1$. The integrand $\cos x\over\sqrt{1-x^2}$ is strictly positive for $-1\lt x\lt1$, hence

$$\int_{-1}^1{\cos x\over\sqrt{1-x^2}}dx\gt0$$

If you dismiss this counterexample by requiring $n$ and $m$ to both be positive, then consider $n=1$ and $m=2$. Wolfram Alpha finds

$$\int_{-1}^1{\cos x\cos2x\over\sqrt{1-x^2}}dx\approx0.793481$$

(and an exact value in terms of Bessel functions). It shouldn't be too difficult to give a non-Wolfram proof that this integral is positive, but I don't see any easy way to make it obvious.

It seems likely that the integral in question is never equal to $0$. Wolfram Alpha finds in general that

$$\int_{-1}^1{\cos nx\cos mx\over\sqrt{1-x^2}}dx={1\over2}\pi(J_0(m-n)+J_0(m+n))$$

where $J_0$ is a Bessel function of the first kind. A hand-wavy reason why you don't get $0$ out of this is that, while the limits of integration here, $\pm1$, are "natural" values for the square root function in the denominator, they are not particularly natural for the cosine functions in the numerator, for which rational multiples of $\pi$ are more natural. There might be an actual proof based on properties of Bessel functions; I'd love to see one.

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