Download 12 SUMMARY OF INTEGRATION TECHNIQUES Integration by

SUMMARY OF INTEGRATION TECHNIQUES
Integration by Standard forms:
xn+1
1 (Ax + B)n+1
[f(x)]n+1
n
n
n
dx
=
+
c,
(Ax
+
B)
dx
=
+
c,
f
'(x)
[f(x)]
dx
=
x






n+1
n + 1 + c,
A n+1
for n  –1
1

 x dx = ln | x | + c
for n  –1
for n  –1
1
f '(x)
1


 f(x) dx = ln | f(x) | + c
 Ax + B dx = A ln | Ax + B | + c
1
Ax+B
 f '(x) e f(x) dx = e f(x) + c
 ex dx = ex + c
dx = A eAx+B + c



 e
1 aAx+B
af(x)
ax
x
Ax+B
f(x)
a
dx
=
+
c
a
dx
=
+
c
f
'(x)
a
dx
=






ln a
ln a + c
A ln a
Integration of Trigonometric Functions:

 f (x) sin f(x) dx = –cos f(x) + c
 sin x dx = –cos x + c


 cos x dx = sin x + c
2

 sec x dx = tan x + c

 f (x) cos f(x) dx = sin f(x) + c
2

 f (x) sec f(x) dx = tan f(x) + c
 cosec2 x dx = –cot x + c


 sec x tan x dx = sec x + c
 f (x) cosec2 f(x) dx = –cot f(x) + c


 f (x) sec f(x) tan f(x) dx = sec f(x) + c

 cosec x cot x dx = –cosec x + c
2
2

 tan x dx = 
 sec x – 1 dx = tan x – x + c

 f (x) cosec f(x) cot f(x) dx = –cosec f(x) + c
2
 cot2 x dx = 

 cosec x – 1 dx = – cot x – x + c
Using Double Angle Formulas: 
 cos2 x dx
1 + cos 2x
x sin 2x
dx
=2+ 4 +c
2
1 – cos 2x
x sin 2x
=
dx
=2– 4 +c


2
1
=
 sin (a + b)x + sin (a – b)x dx
2
1
=
 sin (a + b)x – sin (a – b)x dx
2
1
=
 cos (a + b)x + cos (a – b)x dx
2
1
=
–2
 cos (a + b)x – cos (a – b)x dx
1 –1 x
a tan a + c
x
sin–1 a + c ,
|x|<a
1
x–a
x>a
2a ln x + a + c ,
1
a+x
ln
|x|<a
2a a – x + c,

ln (sec x) + c ,
|x|<2
ln (sin x) + c ,
0<x<
–ln (cosec x + cot x) + c ,
0<x<

ln (sec x + tan x) + c ,
|x|<2
uv– 
 u  v dx
=
2

 sin x dx
Using Factor Formulas:

 sin ax cos bx dx

 cos ax sin bx dx
 cos ax cos bx dx


 sin ax sin bx dx
Given in formula list:
1

 x2 + a2 dx =
1

 a2 – x2 dx =
1

 x2 – a2 dx =
1

 a2 – x2 dx =

 tan x dx
=
=

 cot x dx

 cosec x dx =

 sec x dx
=


=

 u v  dx
Order for choosing u: Logarithm, Inverse Trigonometric, Algebraic, Trigonometric, Exponential
Integration by Parts:
12
E.g.
e ln x dx
1
E.g.
1
Let u = ln x  u  = x
Let v  = 1  v = x
e
e 1
= [ x ln x ]1 – 
1 x x dx
e
= [ e ln e ] – 
1 1 dx
e
= e – [ x ]1
=e–[e–1]
=1
E.g. 
 ex sin x dx
x
= ex sin x – 
 e cos x dx
x
= ex sin x – [ ex cos x – 
 e (–sin x) dx ]
x
x
x
2
 e sin x dx = e sin x – e cos x
1 x
x

 e sin x dx = 2 e (sin x – cos x) + c
2
d 2
3 x
Find dx ex and deduce 
 x e dx.
d x2
x2
dx e = 2x e
2
3 x

 x e dx
2
1
2
x2
=2
Let u = x2, v = 2xex
 x 2x e dx
2
1
x2
= 2 [ x2 ex – 
 2x e dx ]
2
2
1
= 2 [ x2 ex – ex ] + c
Let u = sin x, v'= ex  u' = cos x, v = ex
Let u = cos x, v'= ex  u' = –sin x, v = ex
Integration by Splitting the Numerator
4x + 5
E.g. 
 x2 + 2x + 3 dx
f (x)
2x + 2
1
1
=2
 rewrite the integral as 
 x2 + 2x + 3 dx + 
 (x + 1)2 + 2 dx
 f(x) dx + 
 quadratic dx
1
x+1
= 2 ln | x2 + 2x + 3 | +
tan–1
+c
2
2
Integration by Partial Fractions
3x3 + x
3
2
1
E.g. 
 break up into partial fractions
 (x + 1)2(x2 + 1) dx = 
 x + 1 – (x + 1)2 – x2 + 1 dx
2
= 3 ln | x + 1 | + x + 1 – tan–1 x + c
Integration by Substitution
1
1
E.g. Use the substitution x = tan  to solve 
2
0 x + 1 dx
dx
x = tan 

= sec2 
 use this to replace dx
d
When x = 0, tan  = 0   = 0
 to replace the two limits

When x = 1, tan  = 1   = 4
1
1
0 x2 + 1 dx
1
1
/4
=
sec2  d
 substitute the integrand 2
, dx and the two limits
2
0
x +1
tan  + 1
/4
=
0 sec  d
/4
= [ ln | sec  + tan  | ]
0
= ln ( 2 + 1 )
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