Calculus: Integration- Part I

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.

ANTIDERIVATIVE OR INDEFINITE INTEGRAL If

` a

f x

is the derivative of the function F x , that is F. x = f x for all x in the ` a

` a

` a

` a

` a

domain of f, then F x is called the antiderivative or indefinite integral of f x . We write Z f x dx = F x + C ` a

` a

F x = x 2 @ 2,

The antiderivative of a given function is not unique. For example,

G x = x 2 + 5, and ` a

H x = x2 ` a

are all antiderivatives of

` a

f x = 2x ` a

as

b c d b c d b c ` a ` a ` a ` a df f f f f f f 2 f f f f f f f 2 f f f f f f f 2 F. x = G. x = H. x = f x or f x @2 = f x +5 = f x = 2x for all x in the dx dx dx

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 of

` a

` a

` a

F x , G x and H x

are identical except for their vertical position. All their

derivatives will be same for any x = x 0 . All indefinite integrals of f x = 2x are then ` a

expressed by the general form of the antiderivative which is F x = x 2 + C where C is the constant of integration or an arbitrary constant.

` a

The notation Z is used for integration. The symbol Z f x dx is used to denote the ` a

indefinite integral of the function f(x). The function f(x) is called the integrand. So, we can write Z 2x 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.

Example : Find the indefinite integral of the function

f x = sin x ` a

Solution :

F x = @ cos x ` a

We know that the derivative of the function Z sin x dx = @ cos x + C

so we write

Example : Find the indefinite integral of the function

We know that

dx

x = 3x , or 2

As the derivative of F x = 3

xf f f f f f Z x 2 dx = f +C

` a

f 3g df f f f f f f f xf f f f f f f

dx

3

` a

f x = x2

Solution : b c df f f f f f f f 3

is F. x = sin x

` a

1f f f = A 3x 2 = x 2 3

3 ` a xf f f f f f f is F. x = x 2 3

3

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 : Differentiation Formula ` df f f f f f f f

dx

xn = nA xn@1

` a df f f f f f f f

dx

a

x = 1

Corresponding Integration Formula n+1

xf f f f f f f f f f f f f f f f Z x n dx = f +C n+1

Z dx = x + C

For Trigonometric Functions :

Differentiation Formula ` df f f f f f f f

sin x = cos x

dx

` df f f f f f f f

a

cos x = @ sin x

dx

` df f f f f f f f

a

tan x = sec 2 x

dx

` df f f f f f f f

a

cot x = @csc 2 x

dx

` df f f f f f f f

a

sec x = sec x tan x

dx

` df f f f f f f f

a

csc x = @ csc x cot x

dx

a

Corresponding Integration Formula Z cos x dx = sin x + C Z sin x dx = @ cos x + C Z sec 2 x dx = tan x + C Z csc 2 x dx = @ cot x + C Z sec x tan x dx = sec x + C Z csc x cot x dx = @ csc x + C

For Exponential & Logarithmic Functions : Differentiation Formula ` xa df f f f f f f f

dx

e = ex

` xa df f f f f f f f

dx

a = a x ln a

a 1f f f ln x = f x dx ` df f f f f f f f

Corresponding Integration Formula Z e x dx = e x + C x

af f f f f f f f f f f Z a x dx = f + C a>0, a ≠ 1 ln a

1f f f Z f dx = ln |x| + C x

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 x dx = c Z f x dx ` a

` a

Z f x F g x dx = Z f x dx F Z g x dx B ` a

` aC

` a

Example : Evaluate Z x 3 dx Solution : 3+1 xf 1f f f f f f f f f f f f f f f f f Z x 3 dx = f + C = x4 + C

3+1

4

1f f f f f f f f f f f f w w w w w w w dx Example : Evaluate Z p x Solution : 1

1f f f f f f f f f f f Z f w w w w w w w dx px

=Z x

@

1f f f f f 2

dx =

@ fff+ 1

2 xf f f f f f f f f f f f f f f f f f f f f f + 1f f f f @ +1 2

w w w w w w w

C = 2 px + C

` a

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 of the given function.

Some examples are

w w w w w w w w w w w w w w w w w w w w w w w w w w w 2 q Z x x + 5 dx ,

` a2 ln xf f f f f f f f f f f f f f f f f f Z

x

dx ,

Z tan3 x sec 2 x dx .

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.

c4

Example : Evaluate Z 3x x 2 + 1 dx b

Solution : Let t = x 2 + 1 dt f f f f f f f f = 2x [ dx 1f f dt [ x dx = f 2 4 3f f4 Z 3x x 2 + 1 dx = Z f t dt

b

c

2

5 3f 3f f ftf f f f f f f f f5 A f +C = f t + C 10 2 5 b c5 3f f f f f f 2 x +1 + C = f 10

=

Example : Evaluate Z sin 3x dx Solution : Let 3x = t dx =

1f f f dt 3

1f f Z sin 3x dx = Z f sin t dt 3

1f 1f f f f f = @ cos t + C = @ cos 3x + C 3 3

Example : Evaluate Z tan2 2x @ 3 dx `

a

Solution : Let 2x @ 3 = t 1f f f dx = dt 2 ` a 1f f 2 Z tan2 2x @ 3 dx = Z f tan t dt

2

H

I

b c 1f 1f f f f f = Z sec 2 t @ 1 dt = J Z sec 2 t dt @Z dt K 2 2 @ A 1f f f tan t @ t + C 2 B a ` aC 1f f f ` = tan 2x @ 3 @ 2x @ 3 + C 2

=

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 = @ ln |cos x| + C Z cot x dx = ln |sin x| + C Z sec x dx = ln |sec x + tan x| + C Z csc x dx = @ ln|csc x + cot x| + C

dx xf f f f f f f f f f f f f f f f f f f f f f 1f f f f f f f Z f = arctan + C 2 2

a a x +a L M L xf M dx 1f @ af f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f L M Z 2 L M+ C = ln 2 L M 2a x+a x @a L M L af M dx 1f + xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f L M Z 2 M+ = lnL 2 L 2a a @ xM a @x

C

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w wM

L dx f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f 2M q 2 M Z f w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w = lnL Lx + x + a M + C q x2 +

L

a2

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w wM

L M dx f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f 2M q 2 Z f w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w = lnL Lx + x @ a M + C L

q x2 @ a2

dx xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Z f w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w = arcsin + C a

qa 2 @ x 2

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w wM

L 1f 1f f fq 2 f f f 2M q 2 M Z q x 2 + a 2 dx = f x x + a 2 + a 2 lnL Lx + x + a M + C

2

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w

2

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w

L

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w wM

L M 1f 1f f fq 2 f f f 2M q 2 Z q x 2 @ a 2 dx = f x x @ a 2 @ a 2 lnL Lx + x @ a M + C

2

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w Z q a 2 @ x 2 dx

=

2

L

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w 1f 1f xf f f fq 2 f f f f f f x a @ x 2 + a 2 arcsin +

2

2

a

C

dx xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f 1f f f f f f f f f Z f w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w = arcsec + C x q x2 @ a2

a

a

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 d uv = u dv + v du, `

a

or

u dv = d uv @ v du `

a

Integrating both sides we get the following formula for Integration by parts, Z u dv = uv @ Z 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 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. (b) The purpose is to find an integral Z v du which is easier to integrate than the original integral Z u dv .

Example : Evaluate Z x sec 2 x dx Solution : Let u = x so du = dx and dv = sec 2 x dx using hence

then v =Z dv =Z sec 2 x dx = tan x

Z u dv = uv @ Z v du Z x sec 2 x dx = x tan x @Z tan x dx

= x tan x @ @ ln |cos x| + C b

c

= x tan x + ln |cos x| + C

Example : Evaluate Z ln x 2 + 4 dx b

c

Solution : Let u = ln x 2 + 4 b

c

and dv = dx

2x f f f f f f f f f f f f f f f f f f du = f dx 2 x +4

so

then v =Z dv =Z dx = x

Z u dv = uv @ Z v du

using

2

2x f f f f f f f f f f f f f f f f f f Z ln x 2 + 4 dx = x ln x 2 + 4 @ Z f dx 2 b

c

b

c

x +4

F G 8f f f f f f f f f f f f f f f f f f = x ln x 2 + 4 @ Z 2 @ f dx 2 x +4 b

c

b c F G 8f f f f f f f f f f f f f f f f f f = x ln x 2 + 4 @ Z 2 @ f dx 2 x +4 b c xf f f f = x ln x 2 + 4 @ 2x + 4 arctan + C 2