INTEGRATION- PART -I Integration or the process of finding the antiderivative is one of the most important
operations in calculus. It is the opposite process of differentiation. differentiation .
ANTIDERIVATIVE OR INDEFINITE INTEGRAL
`a
If f x
`a
`a `a
is the derivative of the function F x , that is F . . x
`a
f x for all x in the
=
`a
domain of f , then F x is called the antiderivative or indefinite integral of f x . We write
Z f f ` x a dx
=
`a
F x
+
C
`a `a `a `a `a `a
G x
F . . x
x 2 + 5,
and H x
=
=
G .
x 2
=
are
ffffb
all
antiderivatives
f x
of
c ffffb c ffffb c
d 2 x = H . . x = f x or x @ 2 dx
=
d 2 x + 5 dx
=
`a `a F x
The antiderivative of a given function is not unique. For example,
d 2 x dx
=
=
x 2 @ 2 ,
=
2 x
as
2 x for all x in the
domain of f . We note that, all the antiderivatives of the function f differ only by a constant (the derivative of the constant value is always zero). Geometrically, this means that the graphs
`a `a `a
of F x , G x and H x
are identical except for their vertical position. All their
`a
derivatives will be same for any x = x 0 . All indefinite indefinite integrals of f x
`a Z ` a
expressed by the general form of the antiderivative which is F x constant of integration or an arbitrary constant. The notation
Z is used for integration. The symbol
=
2 x are then
x 2 + C where C is the
=
f x dx is used to denote the
indefinite integral of the function f(x). The function f(x) is called the integrand. So, we can write
Z 2 x dx
x 2 + C .
=
The indefinite integral of a function is sometimes called the general antiderivative of the function.
INTEGRATION BY INSPECTION We shall solve the first few questions by method of inspection, that is, comparing the integrand with some known derivatives.
`a `a
Example : Find Find the indefi ndefini nite te integr ntegral al of the functi function on
f x
Solution : We know that that the deri derivati vative ve of the functi function on so we write write
Z sin sin x dx
=@
cos x
+
F x
C
f x
Solution :
fffffb
c
d 3 We know that x dx
fffff fffgf 1ff3 x
d x 3 = 3 x , or dx 3 2
ffff is
x 3 As the deri derivative vative of F x = 3 x 3 x x 2 dx = + C 3
Z
ffff
`a
=
`a
F . . x
=@
`a
Example : Find Find the indefi ndefini nite te integr ntegral al of the functi function on
3
A
=
2
sin sin x
=
cos x
`a
is F. x
=
sin sin x
x 2
=
x 2
=
x2
STANDARD INTEGRATION FORMULAS : COMPARISON WITH DIFFERENTIATIONS In the above cases of integration by inspection, the antiderivatives were found by comparing the integrand with some known derivatives. But it is not always possible to do so, as in most cases the integrand does not match with our known derivatives. So, we need some integration formulas. As integration is the reverse process of differentiation, we can make the first few integration formulas directly from the corresponding derivative formulas. These are given below :
For Polynomial Functions : Dif Differen erenttiationFo nFormula
ffff`
d n x dx
ffff` a
d x dx
a
Corresp esponding Integr egrationFo nFormula
Z x dx n xffffff1fff C Z dx dx x C n + 1
=
n A x
n
n@1
1
=
=
=
+
+
+
For Trigonometric Functions :
Dif Diffferen erentia tiatio tion n For Formula ula
ffff`
d sin sin x dx
ffff`
a a a a a a
d cos x dx
ffff`
d tan x dx
ffff`
d cot x dx
ffff`
d sec x dx
ffff`
cos x
=
d csc x dx
Cor Correspon spondi ding ng Integ ntegrratio ation n For Formula ula
Z cos x dx sin x sin x C Z sin x dx cos x C Z se c x dx tan x tan x C Z csc x dx cot x cot x C Z se c x tan x dx se c x C Z csc x csc x cot x dx csc x csc x C =
=
sin x @ sin
=
sec2 x
=
@ csc
=
sec x tan x
=
@ csc x
+
=@
2
2
x
cot x
+
=
2
+
=@
+
=
=@
+
+
For Exponential & Logarithmic Functions : iff ifferen erentia tiatio tion n For Formula ula
ffff`
a ffff` a ffff` a
d x e dx
=
e x
d x a dx
=
a x ln a
d ln x dx
Corr Correespond pondin ing g Inte ntegrat gratio ion n For Formula ula
Z e dx e C ffffff C a>0 , a Z a a dx aln a ln a Z x1ff dx ln x |x| C x
=
x
+
x
=
ff
1 x
x
=
=
+
+
≠
1
INTEGRATION OF COMBINATION OF FUNCTIONS The following rules of integration for addition, subtraction and scaler multiples of functions can be used. Note that, unlike differentiation, product and quotients of functions are not covered in these (which are to be solved using substitution or integration by parts or by some other methods).
Z c f` xa dx
Z ` a
c f f x dx
=
BZ ` a ` aC f x
Example : Evaluate
Z x x
3
F
g x dx
=
Z f f ` xa dx Z g g ` xa dx F
dx
Solution :
Z x x
ffffffff
ff
x 3 1 1 4 x + C = 3+1 4 +
3
dx
=
Example : Evaluate
C
+
Z p wfw1f xwfwfffdx
Solution :
Z p wfw1f xwfwfffdx Z x x =
@
fff
1 2
dx
=
fffffffffffff C
x
@
1 2
+
ff
1 @ + 2
1
+
1
=
2
p wwwxw
+
C
SUBSTITUTION METHOD: CHANGE OF VARIABLES Till now we have discussed integrals of those functions which are readily obtained from derivatives of some known functions. They could be solved using the standard formulas. But some integrals can not be evaluated directly as no standard formula matches the form
` a w w w w w w w w w w w w w l n x f f f f f Z x x q x 5 dx , Z x ffffdx , 2
of the given function.
Some examples are
2
+
Z tan x sec x dx . 3
2
One way of solving them is to substitute some new variables in the integral and thus transforming it to standard form with the new variable. After choosing the
suitable
variable, we have to rewrite the integral in terms of the new variable, so that one or more of the standard formulas can be used. After the integration process is over, the result is to be written in terms of the original variable.
Z 3 xbx
Example : Evaluate
2
c
4
+
1 dx
Solution : Let Let t = x 2 + 1 dt [ = 2 x dx 1 x dx = dt [ 2
ffff
ff
Z 3 xbx
2
+
c
4
1 dx
=
=
=
Z 32fft dt 4
ff fff ffffb
3 t 5 + C A 2 5 3 x 2 + 1 10
c
=
ffff
3 5 t 10
5 +
C
+
C
Z sin sin 3 x dx
Example : Evaluate Solution : Let 3 x
t 1 dx = dt 3 =
ff
1ff Z sin sin 3 x dx Z sin sin t dt 3 =
@
=
ff
1 cos t + C 3
Z tan `2 x 2
Example : Evaluate
a
@3
=
@
ff
1 cos 3 x 3
+
C
dx
Solution : Let 2 x @ 3 = t 1 dx = dt 2
ff
Z tan `2 x 2
a
@3
=
=
=
dx
=
Z 12fftan t dt 2
bffZ c fHJf Z A ff@ ffB ` a ` aC
1 2
sec 2 t @ 1 dt
=
1 2
1 tan t @ t + C 2 1 tan 2 x @ 3 @ 2 x @ 3 2
Z
sec2 t dt @ dt
+
C
IK
OTHER INTEGRATION FORMULAS Apart from the standard formulas given above, the following formulas are also used in integration (these are obtained by using the substitution method) :
Z tan x dx Z cot x dx Z se c x dx Z csc x dx
x| + C
=
@ ln |cos
=
ln |sin x sin x| + C
=
ln |se c x + tan x tan x| + C
=
csc x @ ln|csc x
+
cot x cot x| + C
Z xfffdxffffaffff a1ffarctan axff C L M dx 1 x a Z xffffff affffff 2 afffflnLL xffffffaffMMf C L M dx afffffxffM 1 f f f f f f f f f f f f f f f f L Z a x 2 a lnL a xMff C wwwwwwwwwwwwwwwMM L dx f f f f f f f f f f f f f f f f q L Z q wwwwwwwwwwwwwww ln x x a 2
+
2
=
+
2
=
2
=
@
2
2
@
@
+
x
a
+
2
@
LL
2
+
q wwwwxwwwwwwwww awwwMM
=
ln x +
=
x arcsin a
2
2
@
2
+
2
2
@
2
+
2
@
2
2
@
2
x
x
+
C
+
=
=
=
2
@ a
2
+
2
@
2
2
@
2
=
2
C
ff C wwwwwwwwwwwwwwww 1ff LL 1ff q x x a a ln x 2 2 wwwwwwwwwwwwwwwww 1ff LL 1ff q x x a a ln x 2 2 wwwwwwwwwwwwwwwww 1ff 1ff q
Z x a dx wwwwwwwwwwwwwwww q Z x a dx wwwwwwwwwwwwwwww q Z a x dx 2 x a x Z ffq fwfwfwwfwfdxwfwwfwfwwfwfwfwwfwfffff a1ff arcsec axff C 2
+
2
Z q wfwfwwfwfwfwdxwfwfwfwwfwfwfwwfffff x a Z q wfwfwwfwfwfwdxwfwfwfwwfwfwfwwfffff a x q wwwwwwwwwwwwwwww 2
+
@
=
2
+
+
2
+
2
@
+
2
+
q wwwwxwwwwwwwwwawwMM q wwwwwwwwwwwwwwwwMM 2
2
x 2 @ a 2
+
a 2 arcsin
+
ff
x a
+
C
+
M
+
C C
+
2
INTEGRATION BY PARTS Another useful integration technique for indefinite integrals which do not fit in the basic formulas is integration by parts. We may consider this method when the integrand is a product of two functions.
When u and v are differentiable functions of x which is the variable of integration, then
`a
d uv
=
u dv
+
v du,
or
u dv
=
`a
d uv
@v
du
Integrating both sides we get the following formula for Integration by parts,
Z u dv
=
Z
uv @ v du
While using this method of integration, the given integral is to be separated into two parts, one part being u and the other part, combined with dx, being dv. For this reason parts. this method is called integration by parts.
(a) We have to first choose dv which must be readily integrable to find v. The u function will be the remaining part of the integrand that will be differentiated to find du.
Z v du
(b) The purpose is to find an integral original integral
Example : Evaluate
which is easier to integrate than the
Z u dv.
Z x x sec x dx 2
Solution : Let u = x so du du = dx
Z Z sec x dx
and and dv = sec2 x dx using hence
then v= dv
=
Z u dv uv Z v du Z x x sec x dx x tan x Z tan x dx =
2
2
=
@
@
=
=
x tan x @
=
x tan x
+
b
c
@ ln |cos x |
+
ln |cos x | + C
C
tan x
Example : Evaluate
Z lnbx
2
+
c
4 dx
Solution :
b c
Let u = ln x 2 + 4 and and dv
=
Z lnbx
dx
Z u dv
using 2
+
so
c
4 dx
t hen v =
du =
ff2f xfffffffdx
x 2 + 4
Z dv Z dx
=
=
x
=
Z
uv @ v du
=
b b b b
c Z ffffffffff c Z F fffffffffGf c Z F fffffffffGf c ff
x ln x 2 + 4
=
x ln x 2 + 4
=
x ln x 2 + 4
=
@
@
2 x 2
x 2 + 4
dx
8 2@ 2 dx x + 4
8 2@ 2 dx x + 4 x x ln x 2 + 4 @ 2 x + 4 arct arctan an + C 2 @
