Calculus 2 Chapter 7 Solution

7.1
Problem 1
Use integration by parts to evaluate the integral.
Z
5x sin x dx = −5x cos x + 5 sin x + C

solution.
Let u = 5x, dv = sin x dx =⇒ du = 5 dx, v = − cos x
Z Z
5x sin x dx = (5x)(− cos x) − (−5 cos x) dx = −5x cos x + 5 sin x + C

Problem 2
Evaluate the following integral : Z 4 √ 56
−6 t dt = −32 ln 4 +
1 3

solution.
√ 1 3
Let u = ln t, dv = −6 t dt =⇒ du = dt, v = −4t 2
t
Z 4 √ 3
! 4
Z 4 
h  3
i 4 4t 2 2 3
−6 t dt = ln t · −4t 2 − − dt = −32 ln 4 + 4 t 2
1 1 1 t 3 1
Z 4 √  
16 2 56
−6 t dt = −32 ln 4 + 4 − = −32 ln 4 +
1 3 3 3

Problem 3
Evaluate the indefinite integral.
x3 9x2 + 1 − ln (9x2 + 1)
Z
x2 arctan (3x) dx = tan−1 (3x) − +C
3 162

solution.
3 x3
Let u = tan−1 (3x), dv = x2 dx =⇒ du = 2 dx, v =
9x + 1 3
x3 x3
Z Z
x2 tan−1 (3x) dx = tan−1 (3x) − dx
3 9x2 + 1
 
2 2 u−1 1
Let u = 9x + 1 =⇒ x = , du = 18x dx =⇒ du = x dx
9 18

x3 x3 x3 x3
Z  
u−1
Z Z
−1 −1 1 −1 1 1
tan (3x) − dx = tan (3x) − du = tan (3x) − 1− du
3 9x2 + 1 3 162 u 3 162 u
x3 u − ln (u) x3 9x2 + 1 − ln (9x2 + 1)
= tan−1 (3x) − +C = tan−1 (3x) − +C
3 162 3 162
1
Problem 4
Evaluate the integral Z
10 ln x2 − 1 dx

Note : Use an upper-case “C” for the constant of integration.

solution.
2x
Let u = ln x2 − 1 , dv = 10 dx =⇒ du = 2


dx, v = 10x
x −1
20x2
Z Z Z  
2

2

2

1
10 ln x − 1 dx = 10x ln x − 1 − dx = 10x ln x − 1 − 20 1− 2 dx
x2 − 1 x −1
Z   
2

1 1 1
= 10x ln x − 1 − 20 1+ − dx
2 x+1 x−1
 
2

1 1
= 10x ln x − 1 − 20 x + ln |x + 1| − ln |x − 1| + C
2 2
10x ln x2 − 1 + 10 ln |x + 1| − 10 ln |x − 1| − 20x + C

Problem 5
Suppose that f (1) = −1, f (4) = −8, f ′ (1) = −2, f ′ (4) = −6, and f ′′ is continuous.
Z 4
Find the value of xf ′′ (x) dx.
1
Ans. −15

solution.
Let u = x, dv = f ′′ (x) dx =⇒ du = dx, v = f ′ (x)
Z 4 Z 4
′′ ′ 4
xf (x) dx = [xf (x)]1 − f ′ (x) dx = [4f ′ (4) − f ′ (1)] − [f (4) − f (1)]
1 1
Z 4
xf ′′ (x) dx = (−24 + 2) − (−8 + 1) = −15
1

2
Problem 6
Find the integral.

e3x
Z
e3x sin (7x) dx = [3 sin (7x) − 7 cos (7x)] + C
58

solution.
1
Let u = e3x , dv = sin (7x) dx =⇒ du = 3e3x dx, v = − cos (7x)
7
Z Z
1 3
I = e3x sin (7x) dx = − e3x cos (7x) + e3x cos (7x) dx + C1
7 7
1
Let U = e3x , dV = cos (7x) dx =⇒ dU = 3e3x dx, V = sin (7x)
7
Z Z
1 3 1 3
e3x cos (7x) dx = e3x sin (7x) − e3x sin (7x) dx + C2 = e3x sin (7x) − I + C2
7 7 7 7
   
1 3x 3 1 3x 3 3 3x 1 3x 9 3
I = − e cos (7x)+ e sin (7x) − I + C2 +C1 = e sin (7x)− e cos (7x)− I+ C2 + C1
7 7 7 7 49 7 49 7
e3x
   
58 3 3x 1 3x 3 3
I = e sin (7x) − e cos (7x) + C2 + C1 = [3 sin (7x) − 7 cos (7x)] + C2 + C1
49 49 7 7 49 7
e3x
 
21 49
⇒I= [3 sin (7x) − 7 cos (7x)] + C2 + C1
58 58 58
| {z }
constant
Z 3x
e
⇒ e3x sin (7x) dx = [3 sin (7x) − 7 cos (7x)] + C
58

3
7.2
Problem 1
Evaluate Z
5x sin x dx = −5x cos x + 5 sin x + C

solution.
Let u = 5x, dv = sin x dx =⇒ du = 5 dx, v = − cos x
Z Z
5x sin x dx = (5x)(− cos x) − (−5 cos x) dx = −5x cos x + 5 sin x + C
√
√ 1 1
Let u = 1 + 2x = u − 1 =⇒ du = √ dx =
2x ⇒ dx ⇒ (u − 1) du = dx
2x u−1
Z
1
Z
u−1
Z
1  √  √
√ dx = du = 1 − du = u − ln |u| + C = 1 + 2x + ln |1 + 2x| + C
1 + 2x u u