integration test suite / Chap 7

[Axel Vogt]

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).

L:= [
#1
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),
#2
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),
#3
Int(x^2*sin(x)^6,x) = 5/48*x^3 - # corrected version, 5/48*x^6 ... is a typo
1/192*(x^2-1/18)*sin(6*x) - 1/576*x*cos(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),
#4
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),

#5
Int(x*cos(x)^4/sin(x)^2,x) =
-x*cos(x)*(1/2*sin(x)+1/sin(x)) +1/4*sin(x)^2 + ln(sin(x)) - 3/4*x^2,
#6
Int(x*sin(x)^3/cos(x)^4,x) =
x*(1/3/cos(x)^3 - 1/cos(x)) - 1/6*sin(x)/cos(x)^2 + 5/6*ln(tan(Pi/4+x/2)),

#7
Int(x*sin(x)/cos(x)^3,x) =
x/2/cos(x)^2 - 1/2*tan(x),
#8
Int(x*sin(x)^3/cos(x),x) =
1/4*x*cos(2*x) - 1/8*sin(2*x) + Int(x*tan(x), x),
#9
Int(x*sin(x)^3/cos(x)^3, x) =
x/2/cos(x)^2 - 1/2*tan(x) - Int(x*tan(x), x),

#10
Int((2*x+sin(2*x))/(x*sin(x)+cos(x))^2,x) =
-2*cos(x)/(x*sin(x)+cos(x)),
#11
Int((x/(x*cos(x)-sin(x)))^2,x) =
(x*sin(x)+cos(x))/(x*cos(x)-sin(x))
]:

[clicl...@freenet.de]

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 "