```Date: Apr 26, 2013 1:44 PM
Author: Axel Vogt
Subject: integration test suite / Chap 7

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 (correctionfound with Maple)In excercise 8 and 9 the author gives an integral, which Maple knows in termsof polylogarithm (a matter of taste how to write it).Maple finds all the solution (have not checked for compact results).L:= [#1Int(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),#2Int(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),#3Int(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),#4Int(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),#5Int(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,#6Int(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)),#7Int(x*sin(x)/cos(x)^3,x) =   x/2/cos(x)^2 - 1/2*tan(x),#8Int(x*sin(x)^3/cos(x),x) =   1/4*x*cos(2*x) - 1/8*sin(2*x) + Int(x*tan(x), x),#9Int(x*sin(x)^3/cos(x)^3, x) =   x/2/cos(x)^2 - 1/2*tan(x) - Int(x*tan(x), x),#10Int((2*x+sin(2*x))/(x*sin(x)+cos(x))^2,x) =   -2*cos(x)/(x*sin(x)+cos(x)),#11Int((x/(x*cos(x)-sin(x)))^2,x) =   (x*sin(x)+cos(x))/(x*cos(x)-sin(x))]:
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