Is this integral evaluation legitimate? I would like to evaluate the following integral:
$$\int_{-1}^{1}\frac{1}{\sqrt{1-x^{2}}}\cos (2\arccos (x))\cos (3\arccos (x)){\mathrm{d} x}$$
Some experience from taking a Numerical Methods course gives me the idea that the integrand can be thought of as some weight function $w(x) = \frac{1}{\sqrt{1-x^{2}}}$ multiplied by some trig terms which appear to be Chebyshev Polynomials of the first kind. Due to the orthogonality of Chebyshev Polynomials with respect to the weight function, I can conclude that the integral evaluates to $0$ because the Chebyshev Polynomials have different indices.
However, I would like to calculate this integral with elementary methods to be sure.
With some manipulation of the trigonometric terms I find that:
$$\cos (2\arccos (x)) = -\cos(2\arcsin(x)) $$
$$\cos (3\arccos (x))= -\sin(3\arcsin(x)) $$
I make the substitution:
$$u = \arcsin(x)$$
$${\mathrm{d} u} = \frac{1}{\sqrt{1-x^{2}}}{\mathrm{d} x}$$
Then the original integral transforms into:
$$\int_{\infty}^{\infty}\cos (2u)\sin (3u){\mathrm{d} u}$$
From which I conclude equals $0$ because:
$$\int_{a}^{a}f(x){\mathrm{d} x} = 0$$
I feel that I have done something ill advised because sine and cosine do not converge to a limit at infinity. This reminds me of the Cauchy Principal Value. Are my manipulations legitimate?

Hint :
Both function $\cos(2\arccos x)$ and $\cos(3\arccos x)$ are Chebyshev polynomials and
$$
\cos(2\arccos x)=2x^2-1 \\
\cos(3\arccos x)=4x^3 -3x
$$

Let $$f(x)=\frac{\cos(2\arccos x)\cos(3\arccos x)}{\sqrt{1-x^2}} \\
=\frac{(2x^2-1)(4x^3 -3x)}{\sqrt{1-x^2}}$$
Then $f\left(\frac{\sqrt{3}}{2}\right) = 0$ and $\forall x \in \left(\frac{\sqrt{3}}{2}, 1\right)$, $f(x)>0$
Since $0<(2x^2 - 1)(4x^3 - 3x)< 1$ on $\left(\frac{\sqrt{3}}{2}, 1\right)$, multiply $\frac{1}{\sqrt{1-x^2}}$ both side and we know 
$$
0<f(x)<\frac{1}{\sqrt{1-x^2}} \quad \quad \left(\frac{\sqrt{3}}{2}<x<1\right)
$$
So improper integral of $f$ at near of $x=1$ is converge. since $f$ is odd, improper integral of $f$ at near of $x=-1$ is also converge. Then our integral is zero.
