1not that much sure
by substituting x/2 = t
12dx =dt
we get formate
∫sin (2n+1)tsin tdt
with upper limit 2Ï€
and lower limit -2Ï€
\hspace{-16}\bf{\int_{-\pi}^{\pi}\frac{\sin (0.5+n)x}{2\sin (0.5)x}dx=}$\\\\\\ Where $\mathbf{n\in\mathbb{N}}$ and $\bf{0.5=\frac{1}{2}}$
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5 Answers
9convert the sin in numerator sin(a+b) den you get 2 functions one is odd periodic whose integral becomes other is even cos function. you can easily integrate that function.
0.5∫(sin(nx)cot(x/2)+cos(nx))dx=∫(f(x)+g(x))dx
f(x) is odd clearly so integral is zero
g(x) is left which is easy to integrate
1
71@Souro , How is f(x) odd? I mean sin x is odd, cot x is odd , so f(x) should be even na?
1we want to make a sequence of integrals, each for different values of n belonging to the set of natural numbers.
for n=k , I[k]= (0.5)∫[sin(n+0.5)x]/[sin(x/2)]dx.
with the upper and lower bounds of the integral being Î and -Î respectively.
As the function to be integrated is an even function (being the quotient of two odd functions), we can change the bounds to Î and 0 respectively by making suitable substitution. only that the numerical value of the integral (definite) gets multiplied by 2.
I[k]=∫[sin(n+0.5)x]/[sin(x/2)]dx , lower bound 0 , upper bound Π.
for k=1 ,
I[ 1] = ∫[sin(3x/2)/[sin(x/2)]dx , with the same bounds
= Π[ to evaluate it expand [sin (3x)/2] , sin x/2 in the denominator cancels out giving a constant integral which is equal to Πand an integral of the form ∫cos x dx which is equal to 0 for the given bounds.]
Observe that,
I[k]-I[k-1]= ∫[sin(n+0.5)x]/[sin(x/2)]dx - ∫[sin(n-0.5)x]/[sin(x/2)]dx
= ∫2cos(nx)dx bounds being 0 to Î
=0
[ to evaluate it use sin C - sin D = 2 [cos (C+D)/2][sin(C-D)/2] ].
so all terms of the sequence I[k] are equal as I[k]=I[k-1]=I[k-2]=.......=I[ 1].
so answer is I[ 1] which is Î .
let me know please if u cant understand it
