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Trigonometric integrals

Previous lecture

sin x cos x= 1

2

sin2x dx= 1

4cos2x+C,

sin

2

x dx= 1

2

(1 cos2x) dx= x

21

4sin2x+C

cos2 x dx=

1

2

(1 + cos 2x) dx=

x

2+

1

4sin 2x+C

Integrals of the form

sin ax cos bx dx,

sin ax sin bx dx,

cos ax cos bx dx, a =b

Two basic identities:

sin(x+y) = sin x cos y+ cos x sin y

cos(x+y) = cos x cos y sin x sin y

(for small x, y sin is increasing, +, cos is decreasing, )

Change y toy

sin(x

y) = sin x cos y

cos x sin y

cos(x y) = cos x cos y+ sin x sin y

2sin x cos y= sin (x+y) + sin (x y)

2cos x cos y= cos (x+y) + cos (x y)

2sin x sin y= cos (x y) cos(x+y)

Application to integrals.

Example

sin2x cos3x dx=

1

2

(sin(2x+ 3x) + sin (2x 3x)) dx=

=1

2

sin5x dx 1

2

sin x dx=

= 110

cos5x+1

2cos x+C.

MATH115 T2 Winter 04, A.Potapov 1 www.math.ualberta.ca/apotapov/math115.htm

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Example

cos3x cos5x dx=

1

2

(cos (3x+ 5x) + cos (3x 5x)) dx=

=1

2

cos8x dx+

1

2

cos2x dx=

=

1

16sin 8x+

1

4sin2x+C.

Example

sin3x sin5x dx=

1

2

(cos (3x 5x) cos (3x+ 5x)) dx=

=1

2

cos2x dx 1

2

cos8x dx=

= 1

4sin2x 1

16sin8x+C.

Integrals of the form

sinm x cosn xdx,

tanm x secn xdx,

cotm x cscn xdx.

Idea use substitution to transform to integral of polynomial

Pk(u)du or

Pk(u)

us ds.

Actual substitution depends on m, n, and the type of the integral.

Typical substitutions: u= sin x, cos x, tan x, sec x, cot x, csc x.

Task: find, which substitution integral can be transformed

P(u)duwithout

1 u2 or similar.

Even or odd powers in

cosmx sin

nx dx

cosm

x sinn

x dx= cosm1

x sinn

x cos xdx

cos xdx= d (sin x), u= sin x, but have to express cosm1 x through sin x.

Ifm= 2k+ 1 (odd) we can: cos2k x=

cos2 xk

=

1 sin2 xk = 1 u2k

Ifm= 2k (even) we can not without

1 sin2 x.

Similarly

cosm

x sinn

x dx= cosm

x sinn1

xsin xdx


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Ifn= 2k+ 1 (odd) we can useu= cos x, ifn= 2k (even) cannot.

sincos rule: if one has odd power, use other for substitution.

1) Odd power at cos x, u= sin x, du= cos x dx, cos2 x= 1 u2

cos

2k+1

x sin

n

x dx=

cos

2k

x sin

n

xcos xdx=

1 u2k

u

n

du

Example: no sin x, but still u= sin x

cos5 x dx=

cos2 x

2cos x dx=

1 u2

2du=

=

1 2u2 +u4

du= u 2

3u3 +

1

5u5 +C=

= sin x

2

3

sin3 x+1

5

sin5 x+C.

2) Odd power at sin x, u= cos x, du= sin x dx, sin2 x= 1 u2

cosm x sin2k+1 x dx=

cosm x sin2k x sin xdx=

um

1 u2

kdu

3) Odd powers at both: use for uone with bigger power:

Example: a) u= cos x

cos3 x sin9 x dx=

cos3 x sin8 xsin xdx=

u3

1 u2

4du=

=

u3

1 4u2 + 6u4 4u6 +u8

du=

= 14

cos4 x+4

6cos6 x 6

8cos8 x+

4

10cos10 x 1

12cos12 x+C

b) u= sin x

cos

3

x sin

9

x dx=

cos

2

x sin

9

x cos xdx=

1 u2

u

9

du=

=

u9 u11

du=

1

10u10 1

12u12 +C=

1

10sin10 x 1

12sin12 x+C

which is simpler?

4) Even powers at both. No good substitution. Transform with double angle formulas (x 2x) to smaller

powers:

sin

2 x=

1

cos2x

2 ,

cos

2 x=

1 + cos 2x

2 ,

sinx

cosx

=

sin2x

2


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Example:

sin4 x cos6 xdx=

(sin x cos x)4 cos2 x=

=

1

16sin4 2x

1 + cos 2x

2

1

2d (2x) = [z = 2x]

= 1

64 sin4 zdz [double angle] + 1

64 sin4 zcos zdz [u= sin z] =

= 1

64

1 cos2z

2

2dz+

1

64

u4du=

= 1

256

1 2cos2z+ cos2 2z

dz+

1

320u5 +C=

= z

256 1

256sin z+

1

512

(1 + cos 4z) dz+

1

320sin5 z+C=

= x

128 1

256sin2x+

x

256+

1

2048sin4z+

1

320sin5 2x+C=

= 3256 x 1256sin2x+ 12048sin8x+ 1320sin5 2x+C.

Even or odd powers in

tanm x sec

n x dx

Ifm= 2k+ 1 is odd, you can rewrite and use the substitution u= cos x

tan2k+1 x secn x dx=

sin2k+1 x

cos2k+1+n xdx=

1 u2ku2k+1+n

du.

Example:

tan x dx=

sin x

cos xdx [u= cos x] =

du

u = ln |cos x| +C

Example:

tan3 x dx=

sin3 x

cos3 xdx= [u= cos x] =

1 u2

u3 du=

=

u3du+

u1du=

1

2u2 + ln |u| +C=1

2sec2 x+ ln |cos x| +C

Other possibility to use substitutions

a) u= tan x, du= sec2 xdx, sec2 x= 1 +u2

so the power at sec x must be even, power at tan x may be any;

b) u= sec x, du= tan x sec xdx, tan2 x= u2 1

So the power at tan xmust be odd, power at sec xmay be any;


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1) Even power at sec x, n= 2k+ 2, u= tan x,

tanm x sec2k+2 x dx=

um

1 +u2

k

du.

Example:

tan4 x sec4 x dx=

u4

1 +u2

du=

u4 +u6

du=

=1

5u5 +

1

7u7 +C=

1

5tan5 x+

1

7tan7 x+C

Example: no tan x, but still u= tan x

sec4 x dx=

sec2 x sec2 x dx=

1 +u2

du=

=u+13 u3 +C= tan x+13tan

3 x+C.

2) Odd power at tan x, m= 2k+ 1, u= sec x,

tan2k+1 x secn x dx=

tan2 x

k

secn1 xtan x sec xdx=

u2 1

k

un1du.

Example:

tan xdx= tan1 x sec0 xdx= (sec x)1 tan x sec xdx= u1du=

= ln |u| +C= ln |sec x| +C= ln |cos x| +C

Other types:

tan2k+2 x dx

Reduction formula

tan2k+2 x dx=

tan2k x sin2 x sec2 x dx=

=

tan2k x

1 cos2 x

sec2 x dx=

tan2k x sec2 x dx [u= tan x] +

+

tan2k x dx=

u2kdu+

tan2k2 x

1 cos2 x

sec2 x dx= ...


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Other types:

seckx dx, k = 1, 3, 5, ...

Denote for brevityJk =

seck x dx

k= 1

J1=

sec xdx=

dxcos x

=

cos xdxcos2 x

= [u= sin x]

=

du

1 u2 =1

2

1

1 +u+

1

1 u

du=

=1

2(ln |1 +u| ln |1 u|) +C=1

2ln

1 +u1 u +C=12ln

1 + sin x1 sin x +C=

=1

2ln

(1 + sin x)2

(1 sin x) (1 + sin x)

+C=1

2ln

(1 + sin x)21 sin2 x

+C= ln (1 + sin x)cos x

+C=

= ln

|sec x+ tan x

|+C

k 3: reduction formula

Jk =

seck x dx=

seck2 x sec2 xdx=

seck2 x (tan x) dx=

= seck2 x tan x

tan x(k 2)seck3 x tan x sec xdx=

= seck2 x tan x

(k

2) tan

2 x seck2 xdx=

= seck2 x tan x (k 2)

sec2 x 1

seck2 xdx=

= seck2 x tan x (k 2) (Jk Jk2)

Reduction formula

Jk = 1

k 1seck2 x tan x+

k 2k 1 Jk2

Application:

J3=1

2sec x tan x+

1

2J1=

1

2sec x tan x+

1

2ln |sec x+ tan x| +C

Integrals ofIk =

csck x dx can be evaluated in a similar way..


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