<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] Use the Chebyshev formulas: $$\int_0^{\frac{\pi}{4}} \frac{( \color{red}{\sin{(5x)} \color{black}{)^2}}}{\sin^2{x}} -\frac{( \color{blue}{\cos{(5x)} \color{black}{)^2}}}{\cos^2{x}} \,{dx}$$
$$=\int_0^{\frac{\pi}{4}}\frac{\left(  \color{red}{ 16 
\sin^5 (x)- 20 \sin^3( x) + 5\sin (x) } \right)^2}{\sin^2{x}} -\frac{   \left(\color{blue}{16 \cos^5 (x) -20 \cos^3(x) + 5 \cos(x) } \right)^2}{\cos^2{x}} \,{dx}$$
$$=\int_0^{\frac{\pi}{4}} {\left( 16 
\sin^4 (x)- 20 \sin^2( x) + 5\right)^2} -{\left(16 \cos^4 (x) -20 \cos^2(x) + 5 \right)^2}\,{dx}$$
$$=\int_0^{\frac{\pi}{4}} {\left( 16 
\sin^4 (x)- 20 \sin^2( x) + 5\right)^2} -{\left(16 \cos^4 (x) -20 \cos^2(x) + 5 \right)^2}\,{dx}$$Multiply it out; when you do, it cleans up rather nicely:
$$
= \int _0 ^{\pi/4} 8 (2 \cos(2 x) + \cos(6 x)) \,dx
$$
$$
=\left. 8\cdot \left( \sin(2x)+\frac{1}{6}\sin(6x) \right)\right|_0^{\pi/4} = 8\cdot \frac{5}{6}=\frac{20}{3}
$$
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