I would like to compute \(\displaystyle{\int_{{{0}}}^{{{2}\pi}}}{\frac{{{{\sin}^{{4}}{\left({4}{x}\right)}}}}{{{{\sin}^{{2}}{\left({x}\right)}}}}}{\left.{d}{x}\right.}\)

Leia Sullivan

Leia Sullivan

Answered question

2022-04-07

I would like to compute
02πsin4(4x)sin2(x)dx

Answer & Explanation

posciad48o

posciad48o

Beginner2022-04-08Added 15 answers

We can compute the integral by Fourier series. Consider
f(x)=sin2(4x)sinx
then the sine series expansion of f is given by
f(x)=sinx+sin3x+sin5x+sin7x
In particular, we see
02πf2(x) dx=02πsin2x+sin2(3x)+sin2(5x)+sin2(7x) dx
since
02πsin(nx)sin(mx) dx=0    if    nm
Lastly, we have that
02πsin2(nx) dx=1202π1cos(2nx) dx=π
Thus, it follows
02πsin4(4x)sin2x dx=4π
Remark: The hard work lies in finding the sine series expansion.
annie996k88v

annie996k88v

Beginner2022-04-09Added 11 answers

One approach: use the double-angle formulas
sin(2a)=2sin(a)cos(a)
and
cos(2a)=12sin2(a)
Specifically, rewrite the numerator as
sin4(4x)=[2sin(2x)cos(2x)]4=[4sin(x)cos(x)(12sin2(x))]4
It's not particularly pretty from there, but at least your denominator term will get fully cancelled

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