mth174
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Find an antiderivative of the function f(z) = z^ 6 + 1 / (cos^2( 9 z) ).
This, by my estimates, seems like an integration by parts:
We let u=z^6, d'u=6z^5; v'= 1/cos^2(9z) v= 1/9tan9(z)
z6/9tan9z-6/9(int(z^5tan9z)
taking the integral, we get u=6z5 u'= 30z^4; v'= tan9z v= 1/9(ln(cos(9z))
6/9z^5(ln(cos(9z))e inte- integral( 30z^4 1/9 ln(cos9z)
taking this integral (factor out 30/9 u=z^4 u'= 4z^3; v'= ln(cos9z) v= 9/ cos9z )
z^4(9/cos9z) - 36integral( z^3/ cos9z)
...I know this is a lot, would I just continue you this on out and then piece everything back together?
I suspect that the integrand is the way you have written it, z^6 + 1 / cos^2(9z).
In this form the integrand has two terms. The first is z^6, with antiderivative z^7/7. The second is 1 / cos^2(9z), with antiderivative 1/9 tan(9z).
The integral would therefore be z^7/7 + 1/9 tan(9z).
Since z^6 is not a factor of the integrand, u = z^6 would not be part of a valid integration by parts.
Mth175
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Okay, on the integral of z^8+1/1+9z^2, I wanted to check and see if my work is right on doing this problem, and so forth. So we have to use long division to reduce this into a polynomial, right, and then we will be left with a remainder, and you said that we would be left with something that would involve using the arctangent function of b/(1+9z^2). When I do the long division, I get 1/9z^6+1/9z^4+1/9z^2+1/9 with a remainder of (8/9)/1+9z^2; if this is correct then the integral of the polynomial is easy; the integral of (8/9)/ (1+9z^2) is fairly easy..which would be the arctan function. I'm sort of confused on where the a*z/ (1+9z^2 comes from, or should be rather.
You started your long division correctly, but I believe you get
1/(9z^6)-1/(9^2 z^4) +1/ (9^3 z^2) -1/9^4 + (9^4 + 1) / 9^4 * 1 / (9 z^2 + 1).
Other then the coefficients everything you say is correct.
The a in my expression would in this case be zero.
You can't reduce this by factoring out anything can you? Like, for instance, factor our z^2: z^2(z^6)+1 / z^2 (9)+1. I saw that being done in book problems and was wondering if it was applicable in this case.