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:

THANK YOUPuji Utomo @cak_puji089672156266