MATH1014 Tutorial 4
MATH1014 Tutorial 4
7.1
Integration by Parts
Corresponding to product rule in differentiation, the formula of integration by parts is
ˆ
ˆ
0
u(x)v (x) dx = u(x)v(x) − v(x)u0 (x) dx,
or a compact form of the formula
ˆ
ˆ
u dv = uv −
v du.
Usually we will choose u(x) to be a function that becomes simpler when differentiated, and v 0 (x) to be a function with
antiderivative immediately available. When working with definite integrals, the formula becomes
ˆ
ˆ
b
u(x)v 0 (x) dx = [u(x)v(x)]ba −
a
b
v(x)u0 (x) dx.
a
ˆ
x tan2 x dx.
1. (J. Stewart, section 7.1, ex.20) Evaluate
Solution
The function tan2 x is easily integrated by rewriting into sec2 x − 1, thus
ˆ
ˆ
x tan2 x dx = x(sec2 x − 1) dx
ˆ
ˆ
2
= x sec x dx − x dx
ˆ
x2
= x d tan x −
2
ˆ
2
x
− tan x dx
= x tan x −
2
x2
= x tan x −
+ ln | cos x| + C.
2
ˆ
√
3
tan−1
2. (J. Stewart, section 7.1, ex.30) Evaluate
1
1
dx.
x
Solution
ˆ
√
3
−1
tan
1
√3 ˆ √3
1
1
−1 1
dx = x tan
−
x d tan−1
x
x 1
x
1
ˆ √3
√
x
−1 1
−1
=
3 tan √ − tan 1 +
dx
2+1
x
3
1
√
ˆ √3
3π π 1
d(x2 + 1)
=
− +
6
4
2 1
x2 + 1
√
√3
3π π 1 =
− +
ln |x2 + 1| 1
4
2
√6
3π π ln 2
=
− +
.
6
4
2
Prepared by Leung Ho Ming
Homepage: http://ihome.ust.hk/~malhm
1
3. (a) For a 6= 0, use integration by parts to derive the reduction formulas
ˆ
ˆ
ˆ
ˆ
xn sin ax n
xn cos ax n
xn cos ax dx =
−
xn−1 sin ax dx and
xn sin ax dx = −
+
xn−1 cos ax dx.
a
a
a
a
ˆ
(b) Evaluate
x3 sin 2x dx.
Solution
(a) Using integration by parts,
ˆ
ˆ
1
x cos ax dx =
xn d sin ax
a
ˆ
xn sin ax 1
sin ax dxn
−
=
a
a
ˆ
xn sin ax n
=
xn−1 sin ax dx,
−
a
a
n
and
ˆ
ˆ
1
xn d cos ax
a
ˆ
xn cos ax 1
=−
+
cos ax dxn
a
a
ˆ
xn cos ax n
=−
+
xn−1 cos ax dx.
a
a
xn sin ax dx = −
(b) By (a),
ˆ
x3 cos 2x
2
x3 cos 2x
=−
2
x3 cos 2x
=−
2
x3 cos 2x
=−
2
x3 sin 2x dx = −
ˆ
ˆ
3
x2 cos 2x dx
2
ˆ
3 x2 sin 2x 2
+
−
x sin 2x dx
2
2
2
ˆ
3 x2 sin 2x
x cos 2x 1
+
− −
+
cos 2x dx
2
2
2
2
3x2 sin 2x 3x cos 2x 3
+
+
− sin 2x + C.
4
4
8
+
1
(1 − x2 )n dx, where n ≥ 1 is a positive integer.
4. Let In denote the integral
0
2n
In−1 .
(a) Show that In =
2n + 1
ˆ 1
22n (n!)2
(b) Show that
, where n! = 1 · 2 · 3 · · · (n − 1) · n.
(1 − x2 )n dx =
(2n + 1)!
0
Solution
(a) Using integration by parts,
ˆ
ˆ
1
(1 − x2 )n dx = [x(1 − x2 )n ]10 −
In =
0
ˆ
1
x d(1 − x2 )n
0
1
2
(−x )(1 − x2 )n−1 dx
= −2n
0
ˆ
1
(1 − x2 − 1)(1 − x2 )n−1 dx
= −2n
0
= −2n(In − In−1 )
= 2nIn−1 − 2nIn .
Hence (2n + 1)In = 2nIn , and In =
2n
In−1 .
2n + 1
2
ˆ
1
(1 − x2 ) dx =
(b) Note that I1 =
0
ˆ
2
. By (a),
3
1
2n
In−1
2n + 1
2n − 2
2n
·
In−2
=
2n + 1 2n − 1
= ···
2n
2n − 2
4
=
·
· · · I1
2n + 1 2n − 1
5
2n − 2
4 2
2n
·
··· ·
=
2n + 1 2n − 1
5 3
2
2
(2n) (2n − 2) · · · (4)2 (2)2
=
(2n + 1)(2n) · · · (5)(4)(3)(2)
22n (n!)2
=
.
(2n + 1)!
(1 − x2 )n dx = In =
0
7.2
Trigonometric Integrals
ˆ
For integrals in the form
ˆ
sinm x cosn x dx, try to convert the integral to
p(u) du where u = sin x or u = cos x, and
p(u) is a sum of powers of u. When this is impossible (m and n are both even), apply identities
sin2 x =
1 + cos 2x
sin 2x
1 − cos 2x
, cos2 x =
, sin x cos x =
2
2
2
to reduce the powers sine and cosine.
ˆ
ˆ
For integrals in the form
tanm x secn x dx, try to convert the integral to
p(v) dv where v = tan x or sec x, and
p(v) is a sum of powers of v.
ˆ
ˆ
ˆ
For integrals in the forms sin mx cos nx dx, sin mx sin nx dx and
cos mx cos nx dx, apply identities
sin(A − B) + sin(A + B)
,
2
cos(A − B) − cos(A + B)
,
sin A sin B =
2
cos(A − B) + cos(A + B)
cos A cos B =
.
2
Any other transformations may also be necessary in dealing with trigonometric integrals.
ˆ π/2
5. (J. Stewart, section 7.2, ex.4) Evaluate
sin5 x dx.
sin A cos B =
0
Solution
ˆ
ˆ
π/2
π/2
5
sin4 x d cos x
sin x dx = −
0
0
ˆ
π/2
(1 − cos2 x)2 d cos x
=−
0
ˆ
0
(1 − 2u2 + u4 ) du
=−
1
1
2
1
= u − u3 + u5
3
5
0
8
=
.
15
The same integral can be calculated using integration by parts (try it yourself).
3
ˆ
π
sin2 t cos4 t dt.
6. (J. Stewart, section 7.2, ex.10) Evaluate
0
Solution
ˆ
ˆ
π
2
π
4
sin t cos t dt =
0
0
ˆ
=
=
=
=
=
1 − cos 2t
·
2
1 + cos 2t
2
2
dt
1 π
(1 − cos2 2t)(1 + cos 2t) dt
8 0
ˆ
1 π
(sin2 2t + sin2 2t cos 2t) dt
8 0
ˆ π
ˆ
1
1 − cos 4t
1 π 2
sin 2t d sin 2t
dt +
8
2
2 0
0
π 3 π !
1
sin 4t
sin 2t
t−
+
16
4
3
0
0
π
.
16
7. (a) (J. Stewart, section 7.2, ex.68) Suppose m and n are positive integers, show that
(
ˆ π
0, m 6= n
.
sin mx sin nx dx =
π, m = n
−π
(b) (J. Stewart, section 7.2, ex.70) A finite Fourier series is given by the sum
f (x) =
N
X
n=1
Show that each coefficient am =
1
π
an sin nx = a1 sin x + a2 sin 2x + · · · + aN sin N x.
ˆ
π
f (x) sin mx dx.
−π
Solution
(a) If m 6= n,
π
ˆ π
ˆ
1 π
1 sin(m − n)x sin(m + n)x
= 0,
−
sin mx sin nx dx =
[cos(m − n)x − cos(m + n)x] dx =
2 −π
2
m−n
m+n
−π
−π
since m + n and m − n are nonzero integers, and sin kπ = 0 for all integers k. If m = n, the integral becomes
π
ˆ π
ˆ
1 π
1
sin 2nx
2
sin nx dx =
(1 − cos 2nx) dx =
= π.
x−
2 −π
2
2n
−π
−π
(b) Multiplying by sin mx and integrating over [−π, π],
ˆ π
ˆ π
f (x) sin mx dx =
−π
N
X
−π
=
N
X
!
an sin nx sin mx dx
n=1
ˆ
π
an
sin nx sin mx dx
−π
n=1
ˆ
π
sin2 mx dx
= am
−π
Thus am =
1
π
ˆ
= am π.
π
f (x) sin mx dx.
−π
Remark The functions {sin nx} and {cos nx} defined on [−π, π] are central in the study of Fourier series. Instead
∞
X
of finite series, one looks for infinite series in the form a0 +
(an cos nx + bn sin nx) to represent functions on
n=1
[−π, π]. These series have wide range of applications in science and engineering.
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