chapter 50: integration using trigonometric substitutions · problems #1 : integration of sin2x...
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Chapter 50:
Integration using
trigonometric substitutions
Learning Materials:
Problems #1 : integration of sin2x ,cos2x ,tan2x and cot2x
Problems #2 : powers of sines and cosines
Problems #3 : products of sines and cosines
Problems #4 : using the sin θ substitution
Problems #5 : using the tan θ substitution
Problems #1 : integration of sin2x ,cos2x,
tan2x and cot2x
Example:
Example:
3 =+
−...
sin1
sin1dx
x
x ( )( )( ) −+
−−= dx
xx
xx
)sin1(sin1
sin1sin1
−
+−= dx
x
xx2
2
sin1
sinsin21
+−
= dxx
xx2
2
cos
sinsin21
+−= dxxxxx 22 tansectan2sec
−+−= dxxxxx 1secsectan2sec 22
−−= dxxxx 1sectan2sec2 2
Cxxx +−−= sec2tan2
xx
xx
xx
22
22
22
csccot1
sec1tan
1cossin
=+
=+
=+
Identity Formulas:
Example:
−= dxx cotdxx cscx cot 222
.dxx cot Hitung 4
4
= dxx cotxcotdxxcot 224Solution :
= dx 1) -x (cscx cot 22
−−= dx 1) -x (cscd(cot x)x cot 22
Cxcot xxcot3
1 3 +++−=
1xcscxcot 22 −=
Example:
.dxx secx tan Hitung 43
5 .dxx secx secx tan 223
=
.d(tan x) x)tan(1x tan 23
+=
.d(tan x)x tanx tan 53
+=
Cxtan6
1x tan
4
1 64 ++=
xx 22 tan1sec +=
Example:
6
Solution :1xsecxtan 22 −=
.dxx secx tan Hitung 53
.dx x secx tan x secx tan 42
=
Cxsec5
1x sec
7
1 57 +−=
= x)d(secx sec 1)-x(sec 42
= x)d(secx sec-xsec 46
.dxx secx tan 53
Problems #2 : powers of sines
and cosines
Problem #2:
Integrals of the type where either m or n is a positive odd integer. dxxcosx sin nm
• If m is a positive odd integer greather than 1. We may then write :
xcos1xsin 22 −=
sin xx cosx sinxcosx sin n1-mnm =
Now, m – 1 is even; hence : sinm-1x can be expressed as some positive integral
power of sin2x. Since
224 x)cos(1xsin −=
2
1 - m
21-m x)cos(1xsin −=
Becomes a sum of powers of cos x, then can be solved with
substitution methods
x cosx sin n1-m
dxxcosx sin nm
.dxxcosx sin Hitung 323
1
= dxsin x xcosx sindxxcosx sin 322323Solution :
= x)d(-cosxcos x)cos-(1 322
−−= x)d(cosx cosxcos 3832
xcos1xsin 22 −=
Cxcos11
3xcos
5
33
11
3
5
++−=
2 .dxxcosx sin Hitung 253
...Selamat mencoba...
Example:
Problem #2: dxxcosx sin nm
Integrals of the type where either m or n is a positive odd integer.
• If n is a positive odd integer greather than 1. We may then write :
xsin1xcos 22 −=
xcosx cosx sinxcosx sin 1 -n mnm =
Now, n – 1 is even; hence : cosn-1x can be expressed as some positive integral
power of cos2x. Since
224 x)sin(1xcos −=
2
1 - m
21-m x)sin(1xcos −=
Becomes a sum of powers of sin x, then can be solved with substitution methods
x cosx sin 1 - mn
dxxcosx sin nm
Example:
.dxxcosx sin Hitung 52
3
= dx x cosxcosx sindxxcosx sin 4252Solution :
224 x)sin1(xcos −=
4 .dxsin
cos Hitung
32
5
x
x
...Selamat mencoba...
= d(sin x) x)sin - (1x sin 222
+= d(sin x) x)sinx 2sin - (1x sin 422
+= d(sin x)x sinx 2sin -x sin 642
Cxsin7
1x sin
5
2 -x sin
3
1 753 ++=
Integrals of the type where both m and n are even integers,
either both positive or one positive and one zero.
• Such integrals may be handled conveniently by use the identities :
2x) cos(12
1xsin 2 −=
2xsin 2
1 x cossin x =
dxxcosx sin nm
2x) cos(12
1xcos2 +=
.dxx sin Hitung 4
5
Example :
.dx x)(sin 22
=
.dx2x)) cos-(12
1( 2
= .dx 2x) cos2x 2cos-(14
1 2
+=
.dx 4x) cos2
1
2
12x 2cos-(1
4
1 ++=
C4xsin 32
12xsin
4
1x
8
3 ++−=
Problem #2: dxxcosx sin nm
Problems #3 : products of sines and
cosines
= dxcosx sin3x1 dx 4xsin sin2x 2
1 +
C+
−−= cos4x4
1cos2x
2
1
2
1
Example:
2 = dxsin5x sin2x dx 7x cos cos3x 2
1 −
C+
−= cos7x7
1sin3x
3
1
2
1
Problems #4 : using the sin θ
substitution
Example:
If the integrand contains a2 – x2, try x = a sin y atau a cos y
Contoh :
...34xx-2)-(x
dx Hitung
2=
−+1
=−+ 22 2)-(x-12)-(x
dx
34xx-2)-(x
dx
Solusi : Soal integral dapat ditulis sebagaiJika diambil substitusi: x – 2 =
sin y, maka dx = cos y dy.
Sehingga, diperoleh:
−
=ysin1ysin
dyy cos
2
= y cosy sin
dyy cos= dyy csc
Cycot y cscln +−= C2x
34xx--1ln
2
+−
−+=
Problems #5 : using the tan θ
substitution
If the integrand contains a2 + x2, try x = a tan y atau a cot y
Contoh :
...dx 1x
1xx Hitung
2
2
=+
++2
Solusi : Ambil substitusi x = tan y, maka dx = sec2y dy, sehingga :
dyy .secsecy
secytany 2
+
=
+
+=
y tan y sec
y) tan y (sec d
C)tan(sec3
22
3
++= yy
1
y
1x2 +
xdyy .sec
y secytan
y) secy(tan
+
+=
Cx)1x(3
22
3
2 +++=
Example:
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