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

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 Axel Vogt Posts: 1,068 Registered: 5/5/07
integration test suite / Chap 7
Posted: Apr 26, 2013 1:44 PM

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))
]:

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