m 4u Integration

! Standard Integrals

! Function & Derivative

! Product – Integrate by Parts

! Partial Fractions

! A, B

! A, Bx + C

! Completing the Square

! Substitution

! Normal

! Trigonometry

! T - Formula

INTEGRALS

*c is a constant § Standard Integral

§

∫

_{x dx}n = c n xn + + + 1 1 _________n≠−1 §

∫

x 1 dx = ln|x|+c

(

)

n b ax

∫

+ dx =

(

)

(

n

)

c a b ax n ₊ + + + 1 1

∫

f_f'₍(_xx₎) dx = ln| f(x)|+c

∫

_ax1_+b dx = ax b c aln| + |+ 1 §

∫

_{e dx}ax _{= e c} a ax ₊ 1

∫

_{a dx}x = c a ax + ln §

∫

+ 2 2 1 a x dx =

[

x+ x +a

]

+c 2 2 ln §

∫

− 2 2 1 a x dx =

[

x+ x −a

]

+c 2 2 ln

∫

2 ₋ 2 1 a x dx = x a c a x a _+     + − ln 2 1

∫

2 ₋ 2 1 x a dx = a x c x a a _+     − + ln 2 1

§

∫

cosax dx = ax c asin + 1 §

∫

sinax dx = ax c a + −1cos §

∫

_sec2ax _{dx = ax c} atan + 1 §

∫

− 2 2 1 x a dx = a c x ₊       −1 sin

∫

− ₋ 2 2 1 x a dx = a c x ₊       −1 cos __OR__ c a x ₊       −_sin−1 §

∫

+ 2 2 1 x a dx = a c x a _+     −1 tan 1

∫

_cosec2ax _{dx = ax c} a + −1cot

∫

secax tan. ax dx = ax c asec + 1

∫

cosecax cot. ax dx = ecax c

a +

−1cos

Integration by special properties – Theorems

*notes, all these are with respect to dx

∫

a x f 0 ( dx) =

∫

− a x a f 0 ( ) dx

∫

− a a f( dxx) =

∫

a x f 0 ( ) 2 dx ___if f(x) is an EVEN function

∫

− a a f( dxx) = 0 _________-__if f(x) is an ODD function

Trigonometric Identities

sin

2

θ + cos

2

θ = 1

! sin

2

θ = 1 – cos

2

θ

! cos

2

_{θ = 1 – sin}

2

_θ

tan

2

θ = sec

2

θ – 1

! sec

2

_{θ – tan}

2

_{θ = 1}

! sec

2

θ = tan

2

θ + 1

1 + cot

2

θ = cosec

2

θ

! 1 = cosec

2

_{θ – cot}

2

_θ

! cot

2

θ = cosec

2

θ – 1

θ θ θ _tan cos sin ₌ θ θ θ _cot sin cos ₌

cos

2

θ =

(

1 cos2θ

)

2 1 +

sin

2

θ =

(

1 cos2θ

)

2 1 ₋

sinθ =

₂ 1 2 t t +

cosθ =

2₂ 1 1 t t + −

tanθ =

₂ 1 1 t −

T-Formula Substitution

      = 2 tan θ t 2 2 _x a −

= Let

x=asinθ 2 2 _a x −

= Let

x=asecθ 2 2 _x a +

= Let

x=atanθ

Trigonometric Substitution

STANDARD INTEGRALS

Basic Form

∫

_{x dx}n = c n xn + + + 1 1 _________n≠−1 ! General Integral

∫

_{x dx}5 = x +c 6 6

∫

1_x dx or

∫

_{x dx}−1 = ln|x|+c ! 1st special function

(

)

n b ax

∫

+ dx =

(

_a

(

_n

)

)

c b ax n + + + + 1 1 ! Linear function

(

)

∫

+ 6 3 2x dx ₌

(

x+

)

_+c 14 3 2 7

∫

f_f'₍(_xx₎) dx = ln| f(x)|+c

! Function and Derivative

∫

_x22₊₁

x

dx = lnx2+1+c

Logarithm and Exponential

∫

1_x dx or

∫

_{x dx}−1 = ln|x|+c ! 1st special function

∫

_ax1_+b dx = ax b c aln + + 1

∫

_3x1₊₂ dx = ln3x+2 +c 3 1

∫

_{e dx}ax _{= e c} a ax ₊ 1

∫

_e3x _{dx = e}3x _+c 3 1

∫

_{a dx}x = c a ax + ln

∫

_{3 dx}x = c x + 3 ln 3

Trigonometric Functions

∫

sinax dx _{= ax c} a + −1cos

∫

sin3x dx _{= −} x _+c 3 3 cos =

∫

sin2x sin. x _dx =

∫

(

1−cos2x sin

)

. x _dx

=

∫

sinx−cos2x.sinx _dx

∫

_sin3x _dx

∫

sin dxx = −cosx

∫

− cos2x sin. x _dx Let u cos= x x dx du sin − = x du dx sin − = ∴ = − x+ x+c 3 cos cos 3

∫

cosax dx = ax c asin + 1

∫

cos3x dx = x+c 3 3 sin =

∫

cos2x cos. x _dx =

∫

(

1−sin2x cos

)

. x _dx

=

∫

cosx−sin2x.cosx _dx

∫

_cos3x _dx

∫

cos dxx = sinx

∫

− sin2 x cos. x Let u sin= x x dx du cos = x du dx cos = ∴ c x x− + 3 sin sin 3 x du x u sin sin . 2 − − =

_∫

3 3 u = x du x u cos cos . 2

∫

− = 3 3 u =

∫

_sec2ax _dx = ax c atan + 1

∫

sec23x _dx = tan3x+c 3 1

∫

tan dxx ₌

_∫

x x cos sin

dx _________Function and Derivative = -lncosx +c

∫

_tan2x _{dx =}

(

)

∫

sec2_x−1 _dx = tanx−x+c

∫

_tan3x _{dx =}

∫

tan2x tan. x _dx =

∫

(

sec2x−1

)

.tanx _dx

=

∫

sec2x.tanx−tanx _{dx_____Let tan x = u}

= x +lncosx +c

2 tan2

Function and Derivative u = f(x) dx x f du = '( )

[

f(x)

]

n.f'(x)

∫

dx =

[

]

c n x f n + + + 1 ) ( 1

∫

f_f'₍(_xx₎) dx = ln f(x) +c ) ( .) ( ' _{x e}f x f

∫

dx _{= e}f(x) _+c

∫

f'(x).cosf(x) dx = sin f(x)+c

∫

_sinnx_.cosx _dx = c n x n + + + 1 sin 1

[

]

∫

₊ 2 ) ( 1 ) ( ' x f x f dx _{= tan}−1 _f(x)+c

∫

_xln1_x dx = lnlnx +c

∫

_sin5x_.cos4x _dx x dx du sin − = x du dx sin − =

=

∫

_sin4_.sinx_.cos4x _dx

=

∫

(

₁−_cos2x

)

2_.sinx_.cos4x _dx

=

∫

(

₁−_2cos2x+_cos4x

)

_.sinx_.cos4x _dx

=

∫

cos4x.sinx−2

∫

cos6x.sinx+

∫

cos8x.sinx _dx

= − x + x− x+c 9 cos 7 cos 2 5 cos5 7 9

∫

_x2 ₋₂+_x3₋₄ x dx ! split the integral

! take out constants

! complete the square

! standard integrals =

∫

(

)

− − + − 4 2 4 2 2 2 2 1 x x x dx =

∫

∫

− − + − − − dx x x dx x x x 4 2 1 4 4 2 2 2 2 1 2 2 =

(

)

∫

+

∫

_{− −} − − − _dx x dx x x x 5 1 1 4 4 2 2 2 2 1 2 2 =

(

)

c x x x x +     + − − − + − − 5 1 5 1 ln 5 2 4 2 ln 2 1 2

∫

uv'dx=uv−

∫

vu'dx

Integrating by parts Let u =

= dx du Let _dx = dv = v

∫

x cos dxx Let u= x 1 = dx du _dx x dv cos = x v=sin = xsinx−

∫

sinxdx = xsins+cosx+c

∫

ln dxx Let u=lnx x dx du ₌ 1 _dx =1 dv x v= =

∫

ln x dx.1 = xln x−

∫

1dx = xlnx−x+c

These integrals have only one step, other integrals have multi-step

∫

ex_{cos dx = I}x Let _u=ex x e dx du ₌ dx x dv cos = x v=sin = ex_sinx−

∫

ex_sinx_dx Let _u=ex x e dx du = x dx dv sin = x v =−cos

= ex_sinx−

[

−ex_cosx+

∫

ex_cosxdx

]

I = ex_sinx+ex_cosx−_I 2I = ex_sinx₊ex_cosx ∴I = ( sin cos ) 2 1 c x e x ex ₊ x ₊

∫

ex_{sin dx = I}x Let _u=ex x e dx du ₌ dx x dv sin = x v=−cos = −ex_cosx+

∫

ex_cosx_dx Let _u=ex x e dx du = x dx dv cos = x v sin=

= −ex_cosx+

[

ex_sinx−

∫

ex_sinxdx

]

I = −ex_cosx+ex_sinx−_I 2I = ex_sinx₋ex_cosx ∴I = ( sin cos ) 2 1 c x e x ex ₋ x ₊

Partial Fractions

1. Split to Standard Integrals 2. A, B

3. A, Bx + C

4. Completing the Square

∫

x_x+₊₁2dx =

∫

+ + + + 1 1 1 1 x x x dx =

∫

+ + 1 1 1 x dx = x+lnx+1+c

∫

_(x−2)(1_x+1)dx **************** ) 1 )( 2 ( 1 + − x x **************** = 1 2+ + − x B x A When: x = –1 x = 2 1 1 1 = )A(x+1)+B(x−2 = ___0___+__–3B = __-3A =

∫

+ + −2 x 1 B x A dx =

∫

∫

+ − − 1 1 3 1 2 1 3 1 x x dx =

[

lnx−2 −lnx+1

]

+c 3 1 = c x x ₊             + − 1 2 ln 3 1

∫

_QP(_(xx₎)______P(x)≥Q(x)

∫

3₂x_x−₊₁2dx ₌

_∫

(

)

+ − + 1 2 3 1 2 ₂1 2 3 x x dx =

∫ ∫

+ − 1 2 3 1 2 3 ₂1 x dx = x− ln2x+1+c 2 7 2 3

∫

2x2_x−₋x₂+1dx ******** LONG DIVISION ********** 3 2 1 2 2 2 + + − − x x x x ____2x2₋4x_{___} _________3x+1 _________3x−6 _____________7 =

∫

(

2x+3

)

dx +

∫

−2 7 x dx = x2+3x+7lnx−2 +c

∫

₍₄₋1_x+₎₍4_x2₊₁₎ x dx x 4 1+ = )_A(_x2 ₊1 _{+ (Bx}+_c)(4−_x) 4 = x 0 = x __17 __1 = 17A= A + 0+ C4 – 0 =

∫

∫

+ + + − 1 4 _x2 C Bx x A dx =

∫

∫

+ + − − − 1 4 1 2 x x x dx = −ln4−x + lnx2+1+c 2 1 r7

Substitution Finding x in terms of u Let u = = ∴ dx du = ∴dx

(

)

∫

12 ₊ 4 ₄ 1 x x dx =

∫

(

4+u

)

x 2 x 1 2 du =

∫

+ 2 4 1 2

u du Changing the Limits

= _            − 2 tan 2 1 2 1 u Let u= x =       − 2 tan 1 u x=12 4 = x u_u=₌2₂ 3 = 12 4 1 2 tan             − x ₌ 3 2 2 1 2 tan _            − u = _      −     ₋ − 2 2 tan 2 3 2 tan 1 1 = 4 3 π π − = 4 3 π π − = 12 π Let u= x x dx du 2 1 = du x dx=2 2 u x= = 12 π

∫

2 ₊₁ 3 x x dx Let 1_u2 _{= x}2₊ x dx du u 2 2 = x udu dx= 1 2 2 _{=u −} x =

∫

2 3 u x x udu =

∫

_{x du}2 =

∫

(

_u2 −1

)

_du = u −u+c 3 3 =

(

x +1

)

x +1− x +1+c 3 1 2 2 2 IMPLICIT DIFFERENTIATION

TrigonometrySubstitution 2 2 _x a − 2 2 _a x − 2 2 _x a + Let x=asinθ Let x=asecθ Let x=atanθ

∫

_{4 x−} 2 x dx Let x=2sinθ θ θ =2cos d dx θ cos 2 = dx dθ =

∫

− θ θ 2 2 sin 4 4 sin 4 θ cos 2c dθ =

(

)

∫

₋ θ₋ θ2_θ 2 sin 1 4 cos sin 8 dθ =

∫

θ θ θ 2 2 cos cos sin 4 dθ = 4

∫

₂1

(

1−cos2θ

)

_dθ = _ − _+c 2 2 sin 2 θ θ = x− +c      − _2sinθcosθ 2 sin 2 1 = x x x _+c       ₋       −       − 2 4 2 2 2 sin 2 2 1 = x−x −x +c      − 2 4 2 sin 2 1 2 θ sin 2 = x θ sin 2 = x       = − 2 sin 1 x θ 2 x θ 2 4 x− Trig Identity

T-Formula _{t =} tan

( )

θ_{2 ________−π <θ <π} 2 1 2 sin t t + = θ 2 2 1 1 cos t t + − = θ 2 1 2 tan t t − = θ       = 2 sec 2 1 ₂ θ θ d dt

( )

2 2 sec 2 θ θ dt d =

( )

2 2 tan 1 2 θ θ + = dt d 2 1 2 t dt d + = θ

∫

2 ₊ 0 _{3 5cos} 4 π θ dθ 2 1 2 t dt d + = θ =

_{( )}

₂ 1 1 ₁ 2 . 5 3 4 2 2 t dt t t + +

∫

+ − = ₂ 2 2 2 .₁ 1 ) 1 ( 5 ) 1 ( 3 8 t dt t t t + + − + +

∫

=

∫

−₂t2dt 8 8 =

∫

−t2 dt 4 4 *** = dt t

∫

₄₋ 2 1 4 = 4 0 2 2 ln . ) 2 ( 2 1 . 4 π             − + t t = 4 0 2 2 ln π             − + t t = ln 3 4 = A(2 – t) + B(2 + t) 2 = t 4 = 0 + 4B B = 1 2 − = t 4 = 4A + 0 A = 1 ***By Partial Fractions =

∫

−t2 dt 4 4 =

∫

∫

+ + −t 2 t 1 2 1

Reduction Formula

I

n

=

∫

₀2cos π x n

_dx

_______

n ≥ 2

Show:

I

n

= (n – 1)I

n-2

– (n – 1)I

n

I

n

=

∫

cosn−1x.cos1x

dx

Let

(

n

)

x dx du x u n n sin . cos 1 cos 2 1 − − = = − − x v x dx dv sin cos = =

Take out constant, and expand

=

[

_x n− _x

]

₋₋

_∫

(

_n−

)

n−2_x 2_x 0 1 _{1cos .sin} cos . sin 2 π

dx

=

₀+

_∫

(

_n−₁

)

_cosn−2_x.

(

₁−_cos2_x

)

_dx

=

(

_n−1

)

_∫

_cosn 2− _x

_dx

–

(

)

∫

n−_{1 cos}nx

_dx

= (n – 1)I

n-2

– (n – 1)I

n Find:

I

6

I

6

=

∫

₀2cos6 π x

dx

I

6

= (6 – 1)I

6-2

– (6 – 1)I

6

I

6

= 5I

4

– 5I

6

6I

6

= 5I

4

6I

6

= 5(3I

2

– 3I

4

)

6I

6

= 15I

2

– 15I

4

I

4

= 3I

2

– 3I

4

4I

4

= 3I

2

I

4

=

4 3

I

2

6I

6

= 15I

2

- 15

×43

I

2

6I

6

=

4 15

I

2

I

6

=

85

I

2

I

₂

=

∫

2 0 2 cos π x

dx

I

2

=

2 0 2 2 sin sin π       ₋ x x

I

2

=

2 1

I

=

5