#$&*
course Mth 174
Section 7.31) 7.3.6 Find the integral Int(sin^3 (x) cos x dw) using integration tables. Mention which formula you use.
Use 4-22, sin x = w. dw = cos x dx
Int(sin^3 (x) cos x dw) = int(w^3 dw) = sin^4(x)/4 + C
2) 7.3.12 Find the integral Int(1/sqrt(16x^2 + 9) dx) using integration tables. Mention which formula you use.
Int(1/sqrt(16x^2 + 9) dx) = int(1/sqrt((4x)^2-3^2) dx)
@& This would be
int(1/sqrt((4x)^2 + 3^2) dx)*@
Use 6-29, int(1/sqrt((4x)^2-3^2) dx) = ln|4x+sqrt(16x^2 + 9)|/4 +C
3) 7.3.15 Find the integral Int(x^4 e^(3x) dx) using integration tables. Mention which formula you use.
Use 3-14, Int(x^4 e^(3x) dx) = e^(3x)x^4/3 - 4x^3 e^(3x)/9 - 12x^2 e^(3x)/27 - 24x e^(3x)/81 - 24 e^(3x)/243 + C
4) 7.3.18 Find the integral Int(u^3 ln(3u) du) using integration tables. Mention which formula you use.
Use 3-13, Int(u^3 ln(3u) du) = x^4ln(3x)/4 - x^4/16 + C
5) 7.3.24 Find the integral Int(e^(3x)(cos 5x) dx) using integration tables. Mention which formula you use.
Use 2-9, Int(e^(3x)(cos 5x) dx)= e^(3x)/(9+25)[3cos(5x) + 5sin(5x)] + C
6) 7.3.30 Find the integral Int(tan^5(x) dx) using integration tables. Mention which formula you use.
Int(tan^5(x) dx) = int(tan^3(x)(sec^2(x) - 1) dx) = int(tan^3(x) du) - int(tan^3(x) dx) = tan^4(x)/4 - int(tan(x)(sec^2(x) - 1) = tan^4(x)/4 - tan^2(x)/2 + int(sin(x)/cos(x) dx) = tan^4(x)/4 - tan^2(x)/2 - ln|cos(x)| + C
7) 7.3.33 Find the integral Int(1/(1 + (z+2)^2) dz) using integration tables. Mention which formula you use.
Use 5-24, Int(1/(1 + (z+2)^2) dz) = arctan (z+2) + C
8) 7.3.36 Find the integral Int(sin^5(x) dx) using integration tables. Mention which formula you use.
Use 4-17, Int(sin^5(x) dx) = -sin^4(x)cos(x)/5 + (4/5)int(sin^3(x) dx) = -sin^4(x)cos(x)/5 + (4/5) -sin^2(x)cos(x)/3 +(4/5)(2/3)int(sin(x) dx) = -sin^4(x)cos(x)/5 + (4/5) -sin^2(x)cos(x)/3 +8/15 -cos(x) + C
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Section 7.4
1) 7.4.6 Integrate the function 2y/(y^3 - y^2 + y - 1) with respect to y.
2y / ( y^3 - y^2 + y - 1) = (ay+b)/(y^2+1)+(cy+d)/(y-1)
ay^2 - ay + by - b + cy^3 + cy + dy^2 + d = 2
a = -1, c = 0, d = b =1
Integrate 2y / ( y^3 - y^2 + y - 1) = int(-1y/(y^2+1) + 1/(y^2+1) + 1/(y-1) dt) = -(1/2)ln|y^2+1| + arctan(y) + ln|y-1| + C
2) 7.4.8 Find Int(15/(16-x^2) dx).
15/(16-x^2) = 15/(x+4)(4-x) = (15/8)/(x+4) + (15/8)/(x-4)
Int((15/8)/(x+4) + (15/8)/(x-4) dx) = (15/8)(ln|x-4| + ln|x+4|) + C
3) 7.4.10 Find Int(9/(y^3 - 9y) dy)
9/(y^3 - 9y) = 1/y - 1/2(y+3) - 1/2(y-3)
Int(9/(y^3 - 9y) dy) = int(1/y - 1/2(y+3) - 1/2(y-3) dx) = ln|y| - ln|y+3|/2 - ln|y-3|/2 + C
@& Note that this could also be expressed as
ln | y | - 1/2 ln | y^2 - 9 | + C*@
4) 7.4.16 Find Int(2/(x^4-x^3) dx) using 2/(x^4-x^3) = A/x + B/x^2 + C/x^3 + D/(x-1).
2/(x^4-x^3) = -2/x - 2/x^2 - 2/x^3 + 2/(x-1)
Int(2/(x^4-x^3) = int(-2/x - 2/x^2 - 2/x^3 + 2/(x-1) dx) = -2ln|x| + 2/x + 1/x^2 + 2ln|x-1| + C
5) 7.4.24 Complete the square and give a substitution which can used to compute Int(1/(x^2 + 2x + 2) dx).
1/(x^2 + 2x + 2) = 1/((x+1)^2 + 1), let (x+1) = tan(z), dz = 1/ cos^2(x) dx
Int(1/(x^2 + 2x + 2) dx) = int(1/(tan(z) + 1)( 1/ cos^2(z) dz) = int(sin^2(z) + cos^2(z) dz) = z + c = arctan(z) - 1 + C
@& Be sure to then rewrite you solution in terms of the original variable.
Your integral gives you z + c, which is equal to arcTan(x+1) + c*@
6) 7.4.28 Find Int((y+2)/(2y^2 + 3y + 1) dy).
(y+2)/(2y^2 + 3y + 1) = 3/(2x+1) - 1/(x+1)
Int((y+2)/(2y^2 + 3y + 1) dy) = int(3/(2x+1) - 1/(x+1) dx) = (3/2)ln|2x+1| - ln|x+1| + C
@& Be careful to express your result in terms of the original variable, which in this case is y.
Your result would be
(3/2)ln|2y+1| - ln|y+1| + C.
Note also that this can be expressed as
1/2 ln ( ( 2 y + 1 )^2 / | y + 1 | ) + C.*@
7) 7.4.29 Find Int((z-1)/sqrt(2z-z^2) dz).
U = 2z-z^2, du = -2(z-1)
Integrate (z-1)/`sqrt(2z-z^2) = int(1/`sqrt(u)/-2 du) = (-1/2)u^(1/2)/(1/2) = -(2z-z^2)^(1/2) + C
8) 7.4.34 Find Int(1/((x+3)(x-5)) dx).
1/((x+3)(x-5) = a/(x+3) + b/(x-5)
Ax - 3a +bx +5b = 1
A=-b, -3a+5b=1
B=-1/8, a=1/8
Int(1/((x+3)(x-5)) dx) = int(1/8(x+3)-1/8(x-5) dx) = ln|x+3|/8 - ln|x-5|/8 + C
9) 7.4.36 Find the partial fraction decomposition of 1/(x * (L-x) ).
1 / (x (L-x)) = a/x + b/(L-x)
1 = aL - ax + bx
aL = 1, b-a=0
a = b = 1/L
1 / (x (L-x)) = 1/Lx + 1/L(L-x)
@& Be careful to use proper signs of grouping
1/(Lx) + 1/(L(L-x)).*@
10) 7.4.38 Find Int(1/(5P^2-5P^3) dP).
(1/(5P^2-5P^3) = a/5p^2 + b/(1-p) = 1/5p + 1/5p^2 + 1/(5(1-p))
d Int(1/(5P^2-5P^3) dP) = int(1/5p + 1/5p^2 + 1/(5(1-p)) dp) = ln|p|/5 - 1/5p - ln|1-p|/5 + C
11) 7.4.44 Find Int(3y^2/(9+y^2) dy).
Let y = 3tan(x), dy/dx = 3/cos^2(x)
Int(3y^2/(9+y^2) dy) = int(3(9tan^2(x))/(9+9tan^2(x) (3/cos^2(x)) dx) = int(9tan^2(x)/(sin^2(x) + cos^2(x)) dx) = 9 int(tan^2(x) dx) = 9 int(sec^2-1) dx) = 9 int(sec^2(x) dx) - 9 int(1 dx) = 9tan(x) - 9x + C = 3y - 9arctan(y/3) + C
12) 7.4.49 Integrate (z - 1) / sqrt( 2 z - z^2) with respect to z.
U = 2z-z^2, du = -2(z-1)
Integrate (z-1)/`sqrt(2z-z^2) = int(1/`sqrt(u)/-2 du) = (-1/2)u^(1/2)/(1/2) = -(2z-z^2)^(1/2) + C
@& Very well done. However do see my notes for a couple of small corrections as well as some alternative forms of solutions. Many of these integrals are of the type my differential equations students are now encountering, and the process of solving those equations requires knowledge of these solution forms.
I look forward to more of your excellent work.*@