Answers to some problems from 7.1
February 10, 2015
In class I recommended a few medium-difficulty problems on integration by parts, from
section 7.1 of the textbook. Some people requested solutions, so here they are:
7.1.13
Z
t sec2 2t dt
Solution. Since this is a polynomial times a non-inverse trig function, it seems like we
probably want to be differentiating t and integrating sec2 2t, especially since the latter
sounds easy to integrate.
So, take u = t and dv = sec2 2t dt. Then du = dt, and
Z
v = sec2 2t dt.
To integrate this, we do a substitution: θ = 2t. Then dt = dθ/2, so
Z
Z
1
1
sec2 θ dθ
2
= tan θ = tan 2t.
v = sec 2t dt =
2
2
2
So we have
dv = sec2 2t dt
1
v = tan 2t
2
u=t
du = dt
Thus
Z
Z
2
t sec 2t dt =
Z
u dv = uv −
1
v du = t tan 2t −
2
t tan 2t 1
− ln |sec 2t| + C,
2
4
R
where we have used the fact that tan x dx = ln | sec x|.
=
1
Z
1
tan 2t dt
2
7.1.24
Z
1
(x2 + 1)e−x dx
0
Solution. We use integration by parts, letting u = (x2 + 1) and dv = e−x dx. We take
v = −e−x , so
dv = e−x dx
v = −e−x
u = (x2 + 1)
du = 2x dx
Thus
Z
1
−x
2
(x + 1)e
0
1
dx = (x2 + 1)(−e−x ) 0 +
Z
1
2x · e−x dx
0
Z 1
2
−1
2
0
= −(1 + 1)e + (0 + 1)e +
2x · e−x dx
0
Z 1
2x · e−x dx.
= −2/e + 1 +
0
R1
To evaluate the integral 0 2x · e−x , we use integration by parts again. This time, we
take u = 2x, and dv = e−x . So
dv = e−x dx
v = −e−x
u = 2x
du = 2 dx
Thus
Z
1
2x · e
−x
0
Z
1
dx = 2x(−e−x ) 0 +
−1
= −2e
1
1 1
2e−x dx = −2xe−x 0 + −2e−x 0
0
−1
− 0 + −2e
+ 2 = −4/e + 2.
So putting everything together,
Z 1
(x2 + 1)e−x dx = −2/e + 1 − 4/e + 2 = −6/e + 3.
0
7.1.26
9
Z
4
2
ln y
√ dy
y
Solution. We use integration by parts, with u = ln y and dv =
u = ln y
du =
dy
y
dy
√ .
y
Thus
dy
dv = √
y
√
v = 2 y,
and so
Z
4
9
Z 9
√
√
√
y
2
dy = 2 9 ln 9 − 2 4 ln 4 −
2
√ dy
y
y
4
4
√ 9
= 6 ln 9 − 4 ln 4 − [4 y]4 = 6 ln 9 − 4 ln 4 + 8 − 12
= 6 ln 9 − 4 ln 4 − 4
dy
√
ln y √ = [2 y ln y]94 −
y
Z
7.1.30
9
√
Z
3
tan−1 (1/x) dx
1
Solution. We do integration by parts, using u = tan−1 (1/x) and dv = dx. So v = x
and
−1
−1
1
dx = 2
dx.
du =
2
2
1 + (1/x) x
x +1
So
u = tan−1 (1/x)
−1
du =
1 + x2
Thus
Z
−1
dv = dx
v=x
−1
tan (1/x) dx = x tan (1/x) +
Z
x dx
.
1 + x2
Making the substitution t = 1 + x2 , so that dt = 2x dx, the latter integral becomes
Z
Z
x dx
dt
1
1
=
= ln t + C = ln(1 + x2 ) + C.
2
1+x
2t
2
2
Therefore, the original integral becomes
Z
Z
x dx
1
−1
−1
tan (1/x) dx = x tan (1/x) +
= x tan−1 (1/x) + ln(1 + x2 ) + C.
2
1+x
2
3
Thus, the definite integral becomes
√
Z
1
7.1.33
3
√3
1
tan−1 (1/x) dx = x tan−1 (1/x) + ln(1 + x2 )
2
1
√
1
1
1
+ ln(1 + 3) − 1 tan−1 (1) − ln(1 + 12 )
= 3 tan−1 √
2
2
3
√ π ln 4 π ln 2
= 3 +
− −
.
6
2
4
2
Z
cos x ln(sin x) dx
Solution. We do integration by parts, with u = ln(sin x) and dv = cos x dx. Then
u = ln(sin x)
cos x
du =
dx
sin x
Thus
Z
dv = cos x dx
v = sin x
Z
ln(sin x) cos x dx = sin x ln(sin x) −
sin x
cos x
dx
sin x
Z
= sin x ln(sin x) −
4
cos x dx = sin x ln(sin x) − sin x + C.