Any special tricks for this integral?

  • Thread starter georgeguy
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Main Question or Discussion Point

I think you can use integration by parts to do the integral below.
But, it's going to take a lot time to do it that way.

Does anyone know any special tricks to simplify the integrand and
then evaluate the integral?


integral of sin(101x)*(sinx)^99 dx
 

Answers and Replies

  • #2
Gib Z
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Perhaps replacing the sine's with their exponential definition will help. Welcome to PF by the way.
 
  • #3
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[tex]\frac{\left(e^{-i x}-e^{i x}\right)^{99} \left(e^{-101 i
x}-e^{101 i x}\right)}{1267650600228229401496703205376}[/tex]

how does that help?

incidentally the integral is

[tex]\frac{1}{100} \sin ^{100}(x) \sin (100 x)[/tex]

interesting the exponent goes up by one and the argument of the sine goes down by one
 
  • #4
Gib Z
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Sorry, at that moment of time I did not have any paper to work with, so I just said the first thing I may have tried, though I quickly see it does not help. However, I am at home now, so give me a few minutes to solve this through.
 
  • #5
Gib Z
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Ok well since [tex]\sin (x+y) = \sin x \cos y + \cos x \sin y[/tex], [tex]\sin (101x) = \sin (100x) \cos x + \cos (100x) \sin x[/tex], so the integral becomes:

[tex]\int \sin^{99} x \cos x \sin (100x) dx + \int \sin^{100} x \cos (100x) dx[/tex].

Let these integrals be called I_1 and I_2 respectively, with the desired result being the sum.

For the first integral, use integration by parts, with [tex]u= \sin (100x), dv = \sin^{99} \cos x dx[/tex], so that [tex]du = 100\cos (100x) dx, v = \frac{\sin^{100}x}{100}[/tex].

The integration by parts formula yields:

[tex]I_1 = sin (100x) \frac{\sin^{100}x}{100} - \int \frac{\sin^{100}x}{100} 100 \cos (100x) dx = \frac{ \sin^{100}x \cdot \sin (100x) }{100} - \int \sin^{100}x \cos (100x) dx

= \frac{ \sin^{100}x \cdot \sin (100x) }{100} - I_2[/tex].

Hence, [tex]I_1 + I_2 = \frac{ \sin^{100}x \cdot \sin (100x) }{100} + C [/tex].

Good enough?
 
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