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Topic: integration test suite / Chap 7
Replies: 1   Last Post: May 5, 2013 12:29 PM

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 clicliclic@freenet.de Posts: 1,245 Registered: 4/26/08
Re: integration test suite / Chap 7
Posted: May 5, 2013 12:29 PM

Axel Vogt schrieb:
>
> These are the excercises for Chap 7 in Tomfeev's book: p.334, # 1 - # 4,
> p.342/343 #5 - # 9, p.344, # 10, # 11).
>
> Excercise 3 has a typo, the solution should not start with 5/48*x^6 (correction
> found with Maple)
>
> In excercise 8 and 9 the author gives an integral, which Maple knows in terms
> of polylogarithm (a matter of taste how to write it).
>
> Maple finds all the solution (have not checked for compact results).
>

Thanks. I have converted them to Derive too; the file is appended.
Regarding the evaluations, I have made no changes save one: LN in
example 6 has been converted to ATANH. Verification by differentiation
presents no problems. Instead of the non-elementary INT(x*TAN(x), x) in
examples 8 and 9, Derive 6.10 produces an INT(LN(COS(x)), x). Surprise!
Derive 6.10 just cannot do examples 10 and 11.

I think we should wait for more results (and perhaps chapters?) before
updating the performance table.

Martin.

" Timofeev (1948) Ch. 7, examples 1 - 4 (p. 334) ... "

INT(x^2*COS(x)^5,x)=1/200*x*COS(5*x)+(1/80*x^2-1/1000)*SIN(5*x)+~
5/72*x*COS(3*x)+(5/48*x^2-5/216)*SIN(3*x)+5/4*x*COS(x)+(5/8*x^2-~
5/4)*SIN(x)

INT(x^3*SIN(x)^3,x)=1/12*(x^3-2/3*x)*COS(3*x)-1/12*(x^2-2/9)*SIN~
(3*x)-3/4*(x^3-6*x)*COS(x)+9/4*(x^2-2)*SIN(x)

INT(x^2*SIN(x)^6,x)=5/48*x^3-1/192*(x^2-1/18)*SIN(6*x)-1/576*x*C~
OS(6*x)+3/64*(x^2-1/8)*SIN(4*x)+3/128*x*COS(4*x)-15/64*(x^2-1/2)~
*SIN(2*x)-15/64*x*COS(2*x)

INT(x^2*SIN(x)^2*COS(x),x)=1/3*x^2*SIN(x)^3-1/18*x*COS(3*x)+1/54~
*SIN(3*x)+1/2*x*COS(x)-1/2*SIN(x)

" Timofeev (1948) Ch. 7, examples 5 - 9 (p. 342-343) ... "

INT(x*COS(x)^4/SIN(x)^2,x)=-x*COS(x)*(1/2*SIN(x)+1/SIN(x))+1/4*S~
IN(x)^2+LN(SIN(x))-3/4*x^2

INT(x*SIN(x)^3/COS(x)^4,x)=x*(1/(3*COS(x)^3)-1/COS(x))-SIN(x)/(6~
*COS(x)^2)+5/6*ATANH(SIN(x))

INT(x*SIN(x)/COS(x)^3,x)=x/(2*COS(x)^2)-1/2*TAN(x)

INT(x*SIN(x)^3/COS(x),x)=1/4*x*COS(2*x)-1/8*SIN(2*x)+INT(x*TAN(x~
),x)

INT(x*SIN(x)^3/COS(x)^3,x)=x/(2*COS(x)^2)-1/2*TAN(x)-INT(x*TAN(x~
),x)

" Timofeev (1948) Ch. 7, examples 10 - 11 (p. 344) ... "

INT((2*x+SIN(2*x))/(x*SIN(x)+COS(x))^2,x)=-2*COS(x)/(x*SIN(x)+CO~
S(x))

INT((x/(x*COS(x)-SIN(x)))^2,x)=(x*SIN(x)+COS(x))/(x*COS(x)-SIN(x~
))

" ... end of Timofeev Ch. 7 "

Date Subject Author
4/26/13 Axel Vogt
5/5/13 clicliclic@freenet.de