Integral: Solving sin(101x) sin^99(x) dx

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SUMMARY

The integral of sin(101x) sin^99(x) dx can be solved using complex number representations and trigonometric identities. The transformation of sin(101x) into its exponential form, \(\frac{e^{101ix}-e^{-101ix}}{2i}\), along with the identity for sin^99(x) as Im(e^{99ix}), provides a pathway to the solution. Utilizing reduction formulas and trigonometric identities, the integral simplifies to a sum of cosine functions, leading to a final expression involving multiple cosine terms.

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Homework Statement



\int sin(101x) sin^99(x) dx

Homework Equations



Complex Number

The Attempt at a Solution



sin(101x) = \frac{e^{101ix}-e^{-101ix}}{2i}
sin^99(x) = Im(e^{99ix})

Still trying...
 
Last edited by a moderator:
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\sin x=\frac{e^{ix}-e^{-ix}}{2i}
Does that help?

edit: Ah, sorry. Didn't see the mangled tex. just a minute.
 
There is an identity for sin^nx which transforms it into a sum of regular sines. Perhaps that is a place to start.
 
use reduction formulae
try an identity from elementary trigonometry such as
sin(101x)sin(9x)^9=[exp(101 i x)-exp(-101 i x)][exp(9 i x)-exp(-9 i x)]^9/2^10
from which (or otherwise) one may see that
sin(101x)sin(9x)^9=(1/512)(cos(20 x)-9 cos(38 x)+36 cos(56 x)-84 cos(74 x)+126 cos(92 x)-126 cos(110 x)+84 cos(128 x)-36 cos(146 x)+9 cos(164 x)-cos(182 x))
 

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