Trigonmetric integral

After simplification the integrand becomes $\dfrac{π}{2}\cosec \dfrac{πx}{2}\cot \dfrac{πx}{2}$ . Substituting $y$ for $\dfrac{πx}{2}$ , we get the value of the integral as $\cosec \dfrac{π}{6}-\cosec \dfrac{π}{2}=2-1=1$ . Hence $A=1$ and $9A^2=\boxed 9$ .

By taking out the common factors, we get :

$\int_{\frac{1}{3}}^{1} \frac{\pi [ \cos(\frac{2\pi}{3}x) + \cos(\frac{\pi}{3}x) ]}{2\sin(\frac{\pi}{2}x) [ \sin(\frac{2\pi}{3}x) + \sin(\frac{\pi}{3}x) ]} dx$

Now, the integrand can be factorized using these trigonometric identities :

$\sin(\alpha) + \sin(\beta) = 2\sin(\frac{\alpha + \beta}{2})\cos(\frac{\alpha - \beta}{2})$

$\cos(\alpha) + \cos(\beta) = 2\cos(\frac{\alpha + \beta}{2})\cos(\frac{\alpha - \beta}{2})$

such that: $\alpha : \frac{2\pi}{3}x , \beta : \frac{\pi}{3}x \rightarrow \frac{\alpha - \beta}{2} = \frac{\pi}{6}x , \frac{\alpha + \beta}{2} = \frac{\pi}{2}x$

So...

$\int_{\frac{1}{3}}^{1} \frac{\pi [ 2\cos(\frac{\pi}{2}x )\cos( \frac{\pi}{6}x)\ ]}{2\sin(\frac{\pi}{2}x) [ 2\sin(\frac{\pi}{2}x) \cos( \frac{\pi}{6}x) ]} dx$

After simplification :

$\int_{\frac{1}{3}}^{1} \frac{\pi\cos(\frac{\pi}{2}x )}{2\sin^2(\frac{\pi}{2}x) } dx$ $\rightarrow \int_{\frac{1}{3}}^{1} \frac{\pi}{2} \cot(\frac{\pi}{2}x) \csc(\frac{\pi}{2}x) dx$

u-substitution:

u= $\frac{\pi}{2} x$

du= $\frac{\pi}{2} dx$

$\int_{\frac{\pi}{6} x}^{\frac{\pi}{2}} \cot(u) \csc(u) du = -\csc(\frac{\pi}{2}) - [-\csc(\frac{\pi}{6})] = 1$

$9A^2 \rightarrow \boxed{9}$