Any special tricks for this integral?

[georgeguy]

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

 

[Gib Z]

Perhaps replacing the sine's with their exponential definition will help. Welcome to PF by the way.

 

[ice109]

\frac{\left(e^{-i x}-e^{i x}\right)^{99} \left(e^{-101 i<br /> x}-e^{101 i x}\right)}{1267650600228229401496703205376}

how does that help?

incidentally the integral is

\frac{1}{100} \sin ^{100}(x) \sin (100 x)

interesting the exponent goes up by one and the argument of the sine goes down by one

 

[Gib Z]

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.

 

[Gib Z]

Ok well since \sin (x+y) = \sin x \cos y + \cos x \sin y, \sin (101x) = \sin (100x) \cos x + \cos (100x) \sin x, so the integral becomes:

\int \sin^{99} x \cos x \sin (100x) dx + \int \sin^{100} x \cos (100x) dx.

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 u= \sin (100x), dv = \sin^{99} \cos x dx, so that du = 100\cos (100x) dx, v = \frac{\sin^{100}x}{100}.

The integration by parts formula yields:

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 <br /> <br /> = \frac{ \sin^{100}x \cdot \sin (100x) }{100} - I_2.

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

Good enough?