Review Sheet Solutions
1. Integrate the following:
(a)
Z
Z
1
u2 du
sin(4x) cos (4x)dx =
4
1
= u3 + C
12
1
=
cos3 (4x) + C
12
2
for u = cos(4x)
(b)
Z
Z 2
1
sin (x)dx =
(1 − cos(2x)) dx
2
Z
1
1 − 2 cos(2x) + cos2 (2x)dx
=
4
Z
1
1
=
x − sin(2x) +
1 + cos(4x)dx
4
2
1
1
1
1
= x − sin(2x) +
x + sin(4x) + C
4
4
8
4
1
1
3
sin(4x) + C
= x − sin(2x) +
8
4
32
4
(c)
Z
Z
1
sin(x) sin(2x)dx =
cos(3x) − cos(−x)dx
2
11
=
sin(3x) + sin(x) + C
2 3
Z
1
x sin(4x)dx = − x cos(4x) +
4
1
= − x cos(4x) +
4
(d)
1
Z
1
cos(4x)dx
4
1
sin(4x) + C
16
(e)
Z
Z
5
x
Z
x
e sin(2x)dx = e sin(2x) − 2 ex cos(2x)dx
Z
x
x
x
= e sin(2x) − 2 e cos(2x) + 2 e sin(2x)dx
Z
x
x
= e sin(2x) − 2e cos(2x) − 4 ex sin(2x)dx
ex sin(2x)dx = ex sin(2x) − 2ex cos(2x) + C
Z
2
1
ex sin(2x)dx = ex sin(2x) − ex cos(2x) + C
5
5
Z 1
1
2
2
ex sin(2x)dx = e sin(2) − e cos(2) +
5
5
5
0
(f)
Z √
Z √
x2 + 1dx =
V 2 + 1V 0 dt
Z
= U V 0 dt
Z 1
11
1
=
t+
1 + 2 dt
2
t 2
t
Z
1
=
t + t−3 + 1 + t−1 dt
4
1 1 2 1 −2
=
t − t + t + ln |t| + C
4 2
2
1
1
1
= (U + V )2 − (U − V )2 + (U + V )
8
8
4
1
+ ln |U + V | + C
4
1
1
1
= (4U V ) + (U + V ) + ln |U + V | + C
8
4
4
√
√
1 √ 2
1
1
= x x + 1 + (x + x2 + 1) + ln |x + x2 + 1| + C
2
8
4
2
1
1
for x = V =
t−
2
t
(g)
Z
Z
x
dx
(x + 1)(x − 1)(x2 + 1)
x
A
B
Cx + D
=
+
+ 2
2
(x + 1)(x − 1)(x + 1)
x+1 x−1
x +1
2
x = A(x − 1)(x + 1) + B(x + 1)(x2 + 1) + (Cx + D)(x + 1)(x − 1)
x = (A + B + C)x3 + (−A + B + D)x2 + (A + B − C)x + (−A + B − D)
(A + B + C) = 0, (−A + B + D) = 0
(A + B − C) = 1, (−A + B − D) = 0
1
−1
get D = 0, A = B = , C =
4
2
Z
Z
1
1
x
x
=
+
−
dx
x4 − 1)
4(x + 1) 4(x − 1) 2(x2 + 1)
Z
1
1
1
1
= ln |x + 1| + ln |x − 1| −
du
4
4
2
2u
for u = x2 + 1
1
1
1
= ln |x + 1| + ln |x − 1| − ln |u| + C
4
4
4
1
1
1
= ln |x + 1| + ln |x − 1| − ln |x2 + 1| + C
4
4
4
2
1 x − 1
= ln 2
+C
4
x +1
x
dx =
4
x −1
3
(h)
Z
du
u(u + 1)2
B
C
1
A
+
= +
2
u(u + 1)
u u + 1 (u + 1)2
1 = A(u + 1)2 + Bu(u + 1) + Cu
= (A + B)u2 + (2A + B + C) + A
thus A = 1, A + B = 0, 2A + B + C = 0
A = 1, B = −1, C = −1
Z
Z
du
1
1
1
=
−
−
du
u(u + 1)2
u u + 1 (u + 1)2
1
= ln |u| − ln |u + 1| +
+C
u+1
(i)
Z
√
x x − 1dx =
Z
Z
=
√
(u + 1) udu
for u = x − 1
u3/2 + u1/2 du
2
2
= u5/2 + u3/2 + C
5
3
2
2
= (x − 1)5/2 + (x − 1)3/2 + C
5
3
(j)
Z
Z
xdx
1
dx
√
√
=
2
1 − x4
1 − u2
1
= arcsin(u) + C
2
1
= arcsin(x2 ) + C
2
4
for u = x2
(k)
Z
(V + 1)V 0 dt
1
1
√
for x = V =
t−
2
t
V2+1
Z
0
(V + 1)V dt
=
U
Z 1 t − 1 + 1 1 1 + 12 dt
2
t
2
t
=
1
t + 1t
2
Z t2 −1+t t2 +1
dt
2
t
t
1
=
t2 +1
2
t
Z 2
1
t − 1 + t t2 + 1 t dt
=
2
t
t2
t2 + 1
Z 2
1
t +t−1
=
dt
2
t2
Z
1
=
1 + t−1 − t−2 dt
2
1
= (t + ln |t| + t−1 ) + C
2
1
= (U + V + ln |U + V | + U − V ) + C by t = U + V, t−1 = U − V
2
1
= (2U + ln |U + V |) + C
2
√
√
1
= x2 + 1 + ln |x + x2 + 1| + C
2
(x + 1)dx
√
=
x2 + 1
Z
5
(l)
Z
dx
=
2
x + 2x + 3
=
=
=
=
Z
dx
completing the square
(x + 1)2 + 2
Z
1
dx
1
2
(x + 1)2 + 1
Z 2
1
dx
x+1
2
( √2 ) 2 + 1
√ Z
√
x+1
2
du
2
arctan(u) + C
for u = √ =
2
2
u +1
2
2
√
x+1
2
arctan( √ ) + C
2
2
6
2. Solve the following separable differential equations:
(a)
dy
− y 2 sec(x) tan(x) = 0 with initial condition y(0) = 1
dx
dy
dx
dy
y2
Z
dy
y2
1
−
y
= y 2 sec(x) tan(x)
= sec(x) tan(x)dx
Z
= sec(x) tan(x)dx
= sec(x) + C
1
− sec(x) + C
y=0
1
1 = y(0) =
− sec(0) + C
1
=
−1 + C
C −1=1
C=2
1
y=
sec(x) + 2
y=
the solution we missed for y 2 = 0
7
(b)
dy
+ xy + x + y + 1 = 0 with initial condition y(0) = 0
dx
dy
= −(xy + x + y + 1)
dx
dy
= −(x + 1)(y + 1)
dx
dy
= −x − 1dx
y+1
Z
Z
dy
= −x − 1dx
y+1
1
ln |y + 1| = − x2 − x + C
2
− 21 x2 −x+C
y=e
−1
y = −1
missed solution for y + 1 = 0
0 = y(0) = eC − 1
eC = 1 so C = 0
1 2
−x
y = e− 2 x
−1
3. Prove the following:
Z
(a) Let An =
xn ex dx. Show An = xn ex − nAn−1 . Integrate by parts with f = xn ,
g 0 = ex .
Z
(b) Let Dm,n =
xm (ln(x))n dx. Show Dm,n =
1
xm+1 (ln(x))n
m+1
n
− m+1
Dm,n−1 . Integrate
by parts with f = (ln(x))n , g 0 = xm .
4. Use the results from above to integrate the following:
(a)
Z
x3 ex dx = A3
= x3 ex − 3A2
= x3 ex − 3(x2 ex − 2A1 )
= x3 ex − 3x2 ex + 6(xex − A0 )
= x3 ex − 3x2 ex + 6xex − 6ex + C
8
(b)
Z
x100 ln(x)dx = D100,1
1
1 101
x ln(x) −
D100,0
101
101
1 2
1 101
x ln(x) −
=
x101 + C
101
101
=
(c)
Z
(ln(x))2 dx = D0,2
= x(ln(x))2 − 2D0,1
= x(ln(x))2 − 2x ln(x) + 2D0,0
= x(ln(x))2 − 2x ln(x) + 2x + C
5. Please determine if the following integrals exist. If they do, find an estimate for them or
show they are infinite. Notice, these problems are mixed, you will not need to use integral
estimation techniques for all of them.
(a)
Z
0
∞
M Z M
e−x dx
xe dx = lim −xe +
M →∞
0
0
M
= 0 + lim −e−x −x
−x M →∞
0
=1
(b)
Z
∞
0
Z
∞
Z0 ∞
So
0
ex
dx
x
ex
1
≥
x
Zx ∞
ex
1
dx ≥
dx = ∞
x
x
0
ex
dx = ∞
x
9
by ex increasing on 0 ≤ x
Z
∞
(c)
0
v+2
dv Note the function is positive so the integral exists
v5 + 1
v+2
v
1
∼ 5 = 4
5
v +1
v
v
v+2
≤ g(v)
Want 5
v +1
v + 2 ≤ v + 2v = 3v
v5 + 1 ≥ v5
v+2
3v
3
So 5
≤ 5 = 4
v +1
v
v
so we expect finite
where g(v) is only leading terms
when v ≥ 1
when v ≥ 1
v+2≤3
v5 + 1 ≥ 1
v+2
3
So 5
≤ =3
v +1
1
Z
∞
0≤
0
Z
∞
(d)
when 0 ≤ v ≤ 1
when 0 ≤ v ≤ 1
when 0 ≤ v ≤ 1
Z ∞
v+2
v+2
dv
+
dv
5
v5 + 1
0 v +1
1
Z ∞
Z 1
3
3dv +
≤
dv
v4
1
0
1 M
= 3 − lim 3 M →∞ v 1
=3+1=4
v+2
dv =
v5 + 1
Z
1
√
e−x sin( x)dx
0
√
|e−x sin( x)| ≤ e−x
Z ∞
Z ∞
√
−x
e sin( x)dx ≤
e−x dx
0
by |sin(θ)| ≤ 1
by the comparison theorem
0
= lim −e−M + 1 = 1
M →∞
Z
∞
(e)
−∞
x2 + 1
dx = 2
x4 + 1
Z
0
∞
x2 + 1
dx By having an even function. This function is positive
x4 + 1
10
so we know it exists.
x2 + 1
x2
1
∼ 4 = 2 so we expect finite
4
x +1
x
x
x2 + 1
≤ g(x)
Want 4
x +1
x2 + 1 ≤ x2 + x2 = 2x2
x4 + 1 ≥ x 4
x2 + 1
2x2
2
So 4
≤ 4 = 2
x +1
x
x
x2 + 1 ≤ 2
x4 + 1 ≥ 1
x2 + 1
2
So 4
≤ =1
x +1
1
Z
0≤2
0
∞
Z
for x ≥ 1
for x ≥ 1
for 0 ≤ x ≤ 1
for 0 ≤ x ≤ 1
Z ∞ 2
x2 + 1
x +1
dx + 2
dx
4
x4 + 1
0 x +1
1
Z 1
Z ∞
2
≤2
2dx + 2
dx
x2
0
1
−2 ∞
= 4 + lim
M →∞ x 1
=4+2=6
x2 + 1
dx = 2
x4 + 1
for g(x) only leading terms
1
11