<s>[INST] Evaluate $\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$ Evaluate $$\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$$
I tried substitutions like $u=\frac{\pi}{4}-x$, and trig identities like $\cos^2{x}=1-\sin^2{x}$ after getting a common denominator.
$$4\int_0^{\frac{\pi}{4}} \frac{\sin^2{(5x)\cos^2{x}-\cos^2{(5x)}\sin^2{x}}}{\sin^2{(2x)}} \mathop{dx}$$
Where should I go from here?  Any help is appreciated!
 [/INST] Making the problem more general for the antiderivative
$$I_{n,m}=\int\left( \frac{\sin^m{((2n+1)x)}}{\sin^m{(x)}} -\frac{\cos^m{((2+1)x)}}{\cos^m{(x)}} \right)\mathop{dx}$$ using  Chebyshev formulae, just as @integrand answered,
$$\frac{\sin{((2n+1)x)}}{\sin{(x)}}$$ is a polynomial of degree $n$ in $\sin^2(x)$ that is to say a polynomial  of degree $n$ in $\cos^2(x)$ and the same for
$$\frac{\cos{((2n+1)x)}}{\cos{(x)}}$$ So, the integrand is a polynomial of degree $mn$ in $\cos^2(x)$.
Reversing the problem, the integrand is then a linear combination of cosines of even multiple angles and then the integral is linear combination of sines of the same  angles.
If we take the case where $m=2$ as in your case, the result of the integrand would be a linear combination of $\cos(2px)$ with $p=1,2,\cdots,2n-1$.
The table below reports  the expression of the integrand
$$\left(
\begin{array}{cc}
 n & \text{integrand} \\
 1 & 8 \cos (2 x) \\
 2 & 8 (2 \cos (2 x)+\cos (6 x)) \\
 3 & 8 (3 \cos (2 x)+2 \cos (6 x)+\cos (10 x)) \\
 4 & 8 (4 \cos (2 x)+3 \cos (6 x)+2 \cos (10 x)+\cos (14 x)) \\
 5 & 8 (5 \cos (2 x)+4 \cos (6 x)+3 \cos (10 x)+2 \cos (14 x)+\cos (18 x)) \\
 6 & 8 (6 \cos (2 x)+5 \cos (6 x)+4 \cos (10 x)+3 \cos (14 x)+2 \cos (18 x)+\cos
   (22 x)) 
\end{array}
\right)$$ where you can notice interesting patterns in the coefficients. You could easily generate the general expression
Conderning the value of the integrals from $0$ to $\frac \pi 4$, they generate the sequence
$$\left\{4,\frac{20}{3},\frac{152}{15},\frac{456}{35},\frac{5156}{315},\frac{67028
   }{3465},\frac{67952}{3003},\frac{1155184}{45045},\frac{22128676}{765765},\frac
   {22128676}{692835}\right\}$$
For sure, we could do similar things for other integer values of $m$.
</s>