CALCULUS REVIEW PROBLEMS FOR MATH 2065

Students are encouraged to solve the following calculus problems for their integer solutions, to
help remind themselves of material from the course prerequisites.

R6
#1 Find 187
eA if A = B − 2 ln( 14
15
) and B = 5 s4 −5s 12
2 +4 ds. This can be done by first

factoring the denominator of the integrand as a polynomial in the variable s2 .

Solution: Let us write down the denominator as a polynomial in s2 .

12 12
= 2
s4 2
− 5s + 4 (s − 4)(s2 − 1)

We will find the partial fraction decomposition of the integrand. Write

12 A B
= 2 + 2 .
(s2 2
− 4)(s − 1) s −4 s −1

and cross multiply the terms to get,

A(s2 − 1) + B(s2 − 4) = 12.

Taking s2 = 4 gives 3A = 12 and hence, A = 4. Similarly taking s2 = 1 gives

−3B = 12 and hence, B = −4. Therefore we have,
 
12 1 1
=4 −
s4 − 5s2 + 4 (s2 − 4) (s2 − 1)

Then the integral is given by,

Z 6 Z 6 Z 6 
12 1 1
ds = 4 ds − ds
5 s − 5s2 + 4
4 2
5 (s − 4)
2
5 (s − 1)
     6
1 s−2 1 s−1
=4 ln − ln
4 s+2 2 s+1 5
     6
s−2 s−1
= ln − 2 ln
s+2 s+1 5
 2
 6
(s − 2)(s + 1)
= ln
(s + 2)(s − 1)2 5
   
49 · 4 36 · 3
= ln − ln
25 · 8 16 · 7
 
49 · 4 · 7
= ln .
50 · 9 · 3

1
 Substituting this we get,
14
A = B − 2 ln( )
 15 
49 · 4 · 7 14
= ln − 2 ln( )
50 · 9 · 3 15
2
 
49 · 4 · 7 · 15
= ln
50 · 9 · 3 · 142
 
7
= ln .
6

Then,
18 A 18 7
e = · =3
7 7 6

40
R∞ x
#2 Find π
A if A = 1 1+x4
dx. This can be done using the substitution u = x2 .

Solution: Let u = x2 then du = 2xdx and we have

Z ∞ Z ∞
x 1 1
dx = · du
1 1 + x4 1 2 1 + u2
∞
1
= tan−1 (u)
2 1
1 π π 
= −
2 2 4
π
= .
8
Then we have,
40 40 π
A= · = 5.
π π 8
√ R3
39√ 3 A 1
#3 Find 4 2
e if A = 2 x3 −x
dx. This can be done using a partial fraction decomposition.

Solution: Let us write down the partial fraction decomposition of the integrand.
1 1
=
x3 −x x(x − 1)(x + 1)

Let
1 A B C
= + +
x3 − x x x−1 x+1

2
 cross multiplying the terms,
1 = A(x − 1)(x + 1) + Bx(x + 1) + Cx(x − 1)
Put values, x = 0, 1, −1 and solve to get A = −1, B = 21 and C = 12 . Then the integral
is given by
Z 3
1 3 1 1 3 1
Z Z Z 3
1 1
3
dx = + −
2 x −x 2 2 x−1 2 2 x+1 2 x
√ 2  3
x −1
= ln
x
√ !2
8 2
= ln ·√ .
3 3
Then we have, √ √ √
39 3 A 39 3 8 2
√ e = √ · · √ = 13.
4 2 4 2 3 3
p R0
#4 Find −70F ′ ( π/6) if F (x) = x sin(s2 )ds. This can be done using the Fundamental
Theorem of Calculus.

Solution: We have
Z 0 Z x
2
F (x) = sin(s )ds = − sin(s2 )ds
x 0

By Fundamental theorem of Calculus F ′ (x) = − sin(x2 ). Then we have

p
−70F ′ ( π/6) = −70(− sin(π/6)) = 35.

#5 Find
eπ
sin2 (ln(t))
Z
98
dt.
π 1 t

Solution: Let u = ln t, then du = 1t dt. Making this substitution in the integral we get,
Z π
98 e sin2 (ln t) 98 π 2
Z
dt = sin (u)du
π 1 t π 0
98 π 1 − cos 2u
Z
= du
π 0 2
  π
49 sin 2u
= u−
π 2 0
= 49

3
#6 Find Z 3
x5
  
18 1 26
− A + ln , if A = 3
dx.
19 3 7 2 1−x
This can be done by first writing x5 = x2 (x3 − 1) + x2 .

Solution: Writing x5 = x2 (x3 − 1) + x2 we have,

Z 3 Z 3 2 3
x5 x (x − 1) + x2
3
dx = dx
2 1−x 2 1 − x3
Z 3
x2
= −x2 + dx
2 1 − x3
3 Z 3
−x3 x2
= + 3
dx
3 2 2 1−x
Z 3
−19 x2
= + 3
dx
3 2 1−x

For the second integral let u = x3 then du = 3x2 dx. Substituting this back we get,
Z 27
−19 1
A= + du
3 8 3(1 − u)
−19 1
= + (− ln |1 − u|) |27
8
3 3  
19 1 26
= − − ln
3 3 7
Therefore we have,
  
18 1 26 18 19
− A + ln =− ·− =6
19 3 7 19 3

#7 Find Z π/2
28 cos(s)
B, if B = ds.
π 0 1 + sin2 (s)
This can be done using the substitution u = sin(s).

Solution: Let u = sin(s), du = cos(s)ds. The integral B is given by,

Z π/2 Z 1
cos(s) 1
2 ds = 2
du
0 1 + sin (s) 0 1+u
= tan−1 (u) |10
π
=
4
Then we have,
28 28 π
B= · = 7.
π π 4

4
#8 Find Z π/3
192
C, if C = sin3 (s)ds.
5 0
3
This can be done by first writing sin (s) = sin(s)(1 − cos2 (s)).

Solution: Let us write sin3 (s) = sin(s)(1 − cos2 (s)) and make the substitution u =
cos(s) then du = − sin(s)ds and we have,
Z π/3 Z π/3
3
sin (s)ds = sin(s)(1 − cos2 (s))ds
0 0
Z 1/2
= −(1 − u2 )du
1
Z 1
= (1 − u2 )du
1/2
 1
u3

= u−
3 1/2
  
1 1 1
= 1− − −
3 2 24
5
=
24
Then we have,
192 192 5
C= · =8
5 5 24

#9 Find
∞
 π  X sink (2s)
18 ln D + 1 , if D(s) = .
12 k=1
k!

This can be done by relating the series to the Maclaurin expansion for f (x) = ex .

Solution: The Maclaurin expansion for f (x) = ex is given by

∞
x
X xk
e = .
k=0
k!

Using this expansion we get,

∞ ∞
X sink (2s) X sink (2s)
D(s) = = − 1 = esin(2s) − 1.
k=1
k! k=0
k!

5
 Then we have,
 π   π 
18 ln D + 1 = 18 ln D +1
12  12
sin( π6 )
= 18 ln e
1
= 18 ·
2
= 9.

#10 Find Z x4
5 ′
√
f (π), if f (x) = sin( s)ds.
2π 3 sin(π 2 ) x2
This can be done using the chain rule and the Fundamental Theorem of Calculus.

Solution: We will first write f (x) as follows,

Z x4
√
f (x) = sin( s)ds
x2
Zx4 √
Z x2 √
= sin( s)ds − sin( s)ds
0 0

Let us denote the first integral by f1 (x) and the second integral by f2 (x). We have by
Fundamental theorem of Calculus and chain rule,
Z g(x)
d
h(s)ds = h(g(x)) · g ′ (x).
dx 0
√
For the first integral, take h(s) = sin( s) and g(x) = x4 . Then,

f1′ (x) = sin(x2 ) · 4x3 .

√
For the second integral, take h(s) = sin( s) and g(x) = x2 . Then,

f2′ (x) = sin(x) · 2x.

Therefore,

f ′ (x) = f1′ (x) − f2′ (x)

= 4x3 sin(x2 ) − 2x sin(x).

Plugging in π gives

f ′ (π) = 4π 3 sin(π 2 ) − 2π sin(π) = 4π 3 sin(π 2 ).

Then we have,
5 ′ 5
f (π) = · 4π 3 sin(π 2 ) = 5 · 2 = 10.
2π 3 sin(π 2 ) 2π 3 sin(π 2 )

6
#11 Find
11(1 + ln2 (2)) π/2
Z
2t sin(t)dt.
2π/2 ln(2) + 1 0

This can be done using two integrations by parts.

Solution: Let u(t) = sin(t) and dv = 2t dt. We then have du = cos(t)dt and v(t) =
2t
ln(2)
. Performing integration by parts once yields,
Z π/2 π/2 Z π/2 t
t 2t sin(t) 2
2 sin(t)dt = − cos(t)dt
0 ln(2) t=0 0 ln(2)
Z π/2
2π/2 sin(π/2) 20 sin(0) 1
= − − 2t cos(t)dt
ln(2) ln(2) ln(2) 0
π/2 Z π/2
2 1
= − 2t cos(t)dt
ln(2) ln(2) 0
We will perform on more integration by parts. This time set u(t) = cos(t) and use the
same v(t) as in the first integration by parts. We may then compute,
Z π/2 Z π/2
t 2π/2 1
2 sin(t)dt = − 2t cos(t)dt
0 ln(2) ln(2) 0
 π/2 Z π/2 !
π/2
 t
2 1 2 cos(t) 1
= − − − 2t sin(t)dt
ln(2) ln(2) ln(2) t=0 ln(2) 0
π/2 Z π/2
2 1 1
= + 2 − 2 2t sin(t)dt
ln(2) ln (2) ln (2) 0
Notice that the same integral now appears in both the LHS and RHS. We can simply
rearrange terms and solve for it now,
 2  Z π/2
ln (2) + 1 t 2π/2 ln(2) + 1
2 sin(t)dt =
ln2 (2) 0 ln2 (2)
Performing a simple division yields the solution to the integral,
Z π/2
1 + 2π/2 ln(2)
2t sin(t)dt = .
0 ln2 (2) + 1
2
However, we still need to multiply by the factor 11(1+ln (2))
2π/2 ln(2)+1
to get our integer answer,

11(1 + ln2 (2)) 1 + 2π/2 ln(2)

  
= 11.
2π/2 ln(2) + 1 ln2 (2) + 1
#12 Find
  Z √√3−1
36 1
√ 2B + ln(2) + π/4 , if B = s arctan(1 + s2 )ds.
3π 2 0

This can be done using the substitution u = 1 + s2 .

7
Solution: We will focus just on B for the time being. Substitute u = s2 + 1 and
du =√ 2sds. Note that the new lower bound is u = 1 and the new upper bound is
u = 3.
Z √3
1
B= arctan(u)du
2 1
We will perform an integration by parts with f (u) = arctan(u), dg = du, g(u) = u, and
df = u2du+1 to obtain,

√
Z √3
1 3 1 u
B = u arctan(u) |u=1 − 2
du
2 2 1 u +1
√
1 √ √  1Z 3 u
= 3 arctan( 3) − arctan(1) − du
2 2 1 u2 + 1
Z √3
π √ 1 u
= (4 3 − 3) − 2
du
24 2 1 u +1

Perform another substitution using v = u2 + 1 and dv = 2udu, which gives a new lower
bound of p = 2 and upper bound of p = 4:
√
3
1 4 dv
Z Z
u
du =
1 u2 + 1 2 2 v
1
= (ln(v)) |4v=2
2
1
= (ln(4) − ln(2))
2
ln(2)
=
2
Plugging this in fully evaluates B,

π  √  ln(2)
B= 4 3−3 − .
24 4
We can now plug our expression for B into the original expression to obtain our integer
solution,
√
      ln(2)  ln(2) 
36 ln(2) 36 π
√ 2B + + π/4 = √ 2 4 3−3 − + + π/4
3π 2 3π 24 4 2
36  π √ 
=√ (4 3 − 3) + π/4
3π 12
√ !
36 π 3
=√ − π/4 + π/4
3π 3
= 12.

8
#13 Find   Z π/2
1
312 D + , if D = sin3 (s) cos2 (s)ds.
160 π/3

This can be done by first writing sin3 (s) = sin(s)(1 − cos2 (s)).

Solution: First write the integral as follows using the identity that cos2 (s) + sin2 (s) =
1,
Z π/2 Z π/2
3 2
sin(s) cos2 (s) 1 − cos2 (s) ds.


sin (s) cos (s)ds =
π/3 π/3

Now perform a substitution with u = cos(s), du = − sin(s)ds, and the new lower bound
of u = 1/2 and upper bound u = 0.
Z π/2 Z 1/2
2 2
u2 (1 − u2 )du


sin(s) cos (s) 1 − cos (s) ds =
π/3 0
Z 1/2
= u2 − u4 du
0
1/2 1/2
u3 u5
   
= −
3 u=0 5 u=0
1 1
= − .
24 160

Plugging D back into our original expression yields,

 
1 1 1 1 312
312(D + ) = 312 − + = = 13.
160 24 160 160 24

#14 Find

√
 Z ln√π/3
42 π
√ 2J + ln(2) + √ + ln( 3/2) , if J = √  sec2 (e2t )e4t dt.
3π 6 3 ln π/6

This can be done using the substitution x = e2t and integration by parts.

Solution: We will first perform the substitution of x = e2t , dx = 2e2t and the new
lower bound of x = π/6 and upper bound x = π/3.
√
Z ln( π/3) Z π/3
2 2t 4t 1
J= √ sec (e )e dt = sec2 (x)xdx
ln( π/6) 2 π/6

9
 We will now perform an integration by parts with u(x) = x, dv = sec2 (x). As a result,
du = dx and v(x) = tan(x),
Z π/3
1
J= x sec2 (x)dx
2 π/6
!
Z π/3
1 π/3
= (x tan(x)) x=π/6
− tan(x)dx
2 π/6
Z π/3 !
1 5π
= √ − tan(x)dx
2 6 3 π/6
5π 1 π/3
= √ − (ln|sec(x)|) x=π/6
12 3 2
5π ln(3)
= √ −
12 3 4

Plugging J into the original expression yields our integer solution,

√ √
     
42 π 42 5π ln(3) π
√ 2J + ln(2) + √ + ln( 3/2) = √ 2 √ − + ln(2) + √ + ln( 3/2)
3π 6 3 3π 12 3 4 6 3
√
 
42 π
=√ √ − ln(3) + ln(2) + ln( 3/2)
3π 3
  √ !
42 π 3
= √
π 3 3
= 14

#15 Find
1
x3 + 1
Z
18 dx.
0 x+1

This can be done by first noting that −1 is a root of x3 + 1.

Solution: We first factor x3 + 1 as x3 + 1 = (x + 1)(x2 − x + 1). The integral then

becomes easy to compute,
Z 1 3 Z 1
x +1
dx = (x2 − x + 1)dx
0 x + 1 0
 3 1  2 1 1
x x
= − + (x)
3 0 2 0 0
1 1
= − +1
3 2
5
=
6

10
 Plugging this into our original expression yields our integer solution,
Z 1 3
x +1 5
18 dx = (18) = 15.
0 x+1 6
#16 Find Z π/2
32
ln(2J + 1), if J = et cos(t)dt.
π 0

This can be done using integration by parts.

Solution: We will first perform an integration by parts calculation on J using u(t) =

cos(t), v(t) = et , du = − sin(t)dt, and dv = et dt:
Z π/2
J= et cos(t)dt
0
π/2 Z π/2
t
et sin(t)dt


= e cos(t) +
0 0
Z π/2
= −1 + et sin(t)dt
0
We will perform another integeration by parts calculation, but this time with u(t) =
sin(t) (the same v).
Z π/2
J = −1 + et sin(t)dt
0
π/2 Z π/2
t
et cos(t)dt


= −1 + e sin(t) −
0 0

= eπ/2 − 1 − J
We can now easily solve for J algebraically,
eπ/2 − 1
J= .
2
Plugging J into our original expression will give us our integer solution,
  π/2  
32 32 e −1
ln(2J + 1) = ln 2 +1
π π 2
32
= ln(eπ/2 )
π
32  π 
=
π 2
= 16.
#17 Find Z 1
136 K 1
e , if K = dx.
9 0 x2 + 5x + 6
This can be done using partial fractions.

11
 Solution: First we factor x2 + 5x + 6 = (x + 3)(x + 2). We can then setup the partial
fraction decomposition of the integrand,
1 1 A B
= = + .
x2 + 5x + 6 (x + 3)(x + 2) (x + 2) (x + 3)

As a result, we have A(x + 3) + B(x + 2) = 1. By combining x’s and constant terms

we get the linear equations A + B = 0 and 3A + 2B = 1. Solving this system equation
yields A = 1 and B = −1. Thus,
1 1 1
= − .
x2 + 5x + 6 x+2 x+3
We can now quickly compute K,
Z 1
1
K= 2
dx
0 x + 5x + 6
Z 1
1 1
= − dx
0 x+2 x+3
1 1
= (ln(x + 2)) − (ln(x + 3))
0 0
= ln(3) − ln(2) − ln(4) + ln(3)
= ln(9/8)

Plugging our computed value of K into the original expression yields the desired integer
value,
136 K 136 ln(9/8) 136
e = e = = 17.
9 9 8
#18 Find
72 π
Z
t sin2 (t)dt.
π2 0
This can be done by integration by parts.

Solution: We will first perform an integeration by parts calculation using u = t, du =

dt, dv = sin2 (t) and v = 2t − sin(t)2cos(t) ,

π 1 π
Z π Z
2 1 2
t sin (t)dt = t − t sin(t) cos(t) − t − sin(t) cos(t)dt
0 2 0 2 0
 2 π
π2 1 π
Z
t
= − + sin(t) cos(t)dt
2 4 0 2 0
π2 π2 1 π
Z
= − + sin(t) cos(t)dt
2 4 2 0
π2 1 π
Z
= + sin(t) cos(t)dt
4 2 0

12
 If we perform a substitution of x = sin(t) and dx = cos(t)dt, we can see that the last
integral is degenerate since both the upper and lower bounds become x = 0. Therefore,
Z π
π2
t sin2 (t)dt = .
0 4
Plugging this into our original expression gives the desired integer solution,
72 π 72 π 2
Z  
2
t sin (t)dt = 2 = 18.
π2 0 π 4

#19 Find
π/6
sin2 (t) sin(2t)
Z
304 2L
e , if L = dt.
17 0 1 + sin4 (t)
This can be done using the substitution u = sin2 (t).

Solution: We will first perform a substitution using u = sin2 (t) and du = 2 sin(t) cos(t)dt =
sin(2t)dt where the bounds become u = 0 and u = 1/4,
Z π/6
sin2 (t) sin(2t)
L= dt
0 1 + sin4 (t)
Z 1/4
u
= du
0 1 + u2
We will perform another substitution, now using v = u2 + 1 and dv = 2udu where the
bounds now become v = 1 and v = 17/16.
Z 1/4
u
L= du
0 1 + u2
1 17/16 1
Z
= dv
2 1 v
1 1
= ln(17/16) − ln(1)
2 2
1
= ln(17/16)
2
Using the computed value for L, we can plug L into the original expression to get the
desired integer solution,
304 2L 304 ln(17/16) 304 · 17
e = e = = 19.
17 17 17 · 16
#20 Find Z π/4
10
(M + 1), if M = ln(1 + tan(s)) sec2 (s)ds.
ln(2) 0

This can be done using the substitution u = 1 + tan(s).

13
Solution: We will first perform a substitution with u = 1 + tan(s), du = sec2 (s) and
the bounds from u = 1 to u = 2 to compute the value of M ,
Z π/4
M= ln(1 + tan(s)) sec2 (s)ds
Z0 2
= ln(u)du
1
2
= (u ln(u) − u)
1
= (2 ln(2) − 2) − (ln(1) − 1)
= 2 ln(2) − 1

Plugging the computed value for M into the original expression yields the desired integer
solution,
10 10
(M + 1) = (2 ln(2) − 1 + 1) = 20.
ln(2) ln(2)

14