Indefinite Integration JEE Notes | EduRev

JEE : Indefinite Integration JEE Notes | EduRev

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J E E - M a t h e m a t i c s
NODE6\ E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#06\Eng\01-INDEFINITE INTEGRATION.p65
If f & F are function of x such that F' (x) = f(x) then the function F is called a PRIMITIVE OR ANTIDERIVATIVE
OR INTEGRAL of f(x) w.r.t. x and is written symbolically as
d
f(x) dx F(x) c { F(x) c} f(x)
dx
? ? ? ? ?
?
, where c is called the constant of integration.
1 . GEOMETRICAL INTERPRETATION OF INDEFINITE INTEGRAL :
f(x)dx F(x) c y(say) ? ? ?
?
, represents a family of curves. The different values of c will correspond to different
members of this family and these members can be obtained by shifting any one of the curves parallel to itself.
This is the geometrical interpretation of indefinite integral.
Let f(x) = 2x. Then 
2
f(x)dx x c. ? ?
?
 For different values                
Y
X X '
Y '
y=x+3
  
2
y=x+2
  
2
y=x+1
  
2
y=x
  
2
y=x–1
  
2
y=x–2
  
2
y=x–3
  
2
P
3
P
2
P
1
P
0
P
– 1
P
–3
P
–2
x = a
of c, we get different integrals. But these integrals are
very similar geometrically.
Thus, y = x
2 
+ c, where c is arbitrary constant, represents
a family of integrals. By assigning different values to c,
we get different members of the family. These together
constitute the indefinite integral. In this case, each integral
represents a parabola with its axis along y-axis.
If the line x = a intersects the parabolas y = x
2
, y = x
2
 +1,
y = x
2
 + 2, y = x
2
 – 1, y = x
2
 – 2 at P
0
, P
1
, P
2
, P
–1
, P
–2
 etc.,
respectively, then 
dy
dx
 at these points equals 2a. This
indicates that the tangents to the curves at these points
are parallel. Thus, 
?
2
2 xdx = x + c = f(x) + c (say),
implies that the tangents to all the curves
f(x) + c, c ? R, at the points of intersection of the
curves by the line x = a, (a ? R) , are parallel.
2 . STANDARD RESULTS :
(i)
n 1
n
(ax b)
(ax b) dx c; n 1
a(n 1)
?
?
? ? ? ? ?
?
?
(ii)
dx 1
n ax b c
ax b a
? ? ?
?
?
?
(iii)
ax b ax b
1
e dx e c
a
? ?
? ?
?
(iv)
px q
px q
1 a
a dx c, (a 0)
p n a
?
?
? ? ?
?
?
(v)
1
sin(ax b)dx cos(ax b) c
a
? ? ? ? ?
?
(vi)
1
cos(ax b)dx sin(ax b) c
a
? ? ? ?
?
(vii)
1
tan(ax b)dx n| sec(ax b)| c
a
? ? ? ?
?
?
(viii)
1
cot(ax b)dx n| sin(ax b)| c
a
? ? ? ?
?
?
(ix)
2
1
sec (ax b)dx tan(ax b) c
a
? ? ? ?
?
(x)
2
1
cos ec (ax b)dx cot(ax b) c
a
? ? ? ? ?
?
(xi)
1
cosec (ax b).cot(ax b)dx cosec (ax b) c
a
? ? ? ? ? ?
?
(xii) sec
?
(ax + b).tan(ax + b)dx = 
1
sec(ax b) c
a
? ?
INDEFINITE INTEGRATION
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NODE6\ E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#06\Eng\01-INDEFINITE INTEGRATION.p65
If f & F are function of x such that F' (x) = f(x) then the function F is called a PRIMITIVE OR ANTIDERIVATIVE
OR INTEGRAL of f(x) w.r.t. x and is written symbolically as
d
f(x) dx F(x) c { F(x) c} f(x)
dx
? ? ? ? ?
?
, where c is called the constant of integration.
1 . GEOMETRICAL INTERPRETATION OF INDEFINITE INTEGRAL :
f(x)dx F(x) c y(say) ? ? ?
?
, represents a family of curves. The different values of c will correspond to different
members of this family and these members can be obtained by shifting any one of the curves parallel to itself.
This is the geometrical interpretation of indefinite integral.
Let f(x) = 2x. Then 
2
f(x)dx x c. ? ?
?
 For different values                
Y
X X '
Y '
y=x+3
  
2
y=x+2
  
2
y=x+1
  
2
y=x
  
2
y=x–1
  
2
y=x–2
  
2
y=x–3
  
2
P
3
P
2
P
1
P
0
P
– 1
P
–3
P
–2
x = a
of c, we get different integrals. But these integrals are
very similar geometrically.
Thus, y = x
2 
+ c, where c is arbitrary constant, represents
a family of integrals. By assigning different values to c,
we get different members of the family. These together
constitute the indefinite integral. In this case, each integral
represents a parabola with its axis along y-axis.
If the line x = a intersects the parabolas y = x
2
, y = x
2
 +1,
y = x
2
 + 2, y = x
2
 – 1, y = x
2
 – 2 at P
0
, P
1
, P
2
, P
–1
, P
–2
 etc.,
respectively, then 
dy
dx
 at these points equals 2a. This
indicates that the tangents to the curves at these points
are parallel. Thus, 
?
2
2 xdx = x + c = f(x) + c (say),
implies that the tangents to all the curves
f(x) + c, c ? R, at the points of intersection of the
curves by the line x = a, (a ? R) , are parallel.
2 . STANDARD RESULTS :
(i)
n 1
n
(ax b)
(ax b) dx c; n 1
a(n 1)
?
?
? ? ? ? ?
?
?
(ii)
dx 1
n ax b c
ax b a
? ? ?
?
?
?
(iii)
ax b ax b
1
e dx e c
a
? ?
? ?
?
(iv)
px q
px q
1 a
a dx c, (a 0)
p n a
?
?
? ? ?
?
?
(v)
1
sin(ax b)dx cos(ax b) c
a
? ? ? ? ?
?
(vi)
1
cos(ax b)dx sin(ax b) c
a
? ? ? ?
?
(vii)
1
tan(ax b)dx n| sec(ax b)| c
a
? ? ? ?
?
?
(viii)
1
cot(ax b)dx n| sin(ax b)| c
a
? ? ? ?
?
?
(ix)
2
1
sec (ax b)dx tan(ax b) c
a
? ? ? ?
?
(x)
2
1
cos ec (ax b)dx cot(ax b) c
a
? ? ? ? ?
?
(xi)
1
cosec (ax b).cot(ax b)dx cosec (ax b) c
a
? ? ? ? ? ?
?
(xii) sec
?
(ax + b).tan(ax + b)dx = 
1
sec(ax b) c
a
? ?
INDEFINITE INTEGRATION
JEEMAIN.GURU
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NODE6\ E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#06\Eng\01-INDEFINITE INTEGRATION.p65
(xiii) sec x dx n sec x tan x c ? ? ?
?
?
x
= n tan c
4 2
? ? ?
? ?
? ?
? ?
?
(xiv) cosec x dx n cosec x cot x c ? ? ?
?
? = 
x
n tan c
2
? ? = n | cosec x cot x| ? ? ? + c
(xv)
1
2 2
dx x
sin c
a
a x
?
? ?
?
? (xvi)
1
2 2
dx 1 x
tan c
a a a x
?
? ?
?
?
(xvii)
1
2 2
dx 1 x
sec c
a a
x x a
?
? ?
?
? (xviii)
2 2
2 2
dx
n x x a
x a
? ?
? ? ?
? ?
? ?
?
?
?
 + c
(xix)
2 2
2 2
dx
n x x a
x a
? ?
? ? ?
? ?
? ?
?
?
?
 + c (xx)
2 2
dx 1 a x
n c
2a a x a x
?
? ?
? ?
?
?
(xxi)
2 2
dx 1 x a
n c
2a x a x a
?
? ?
? ?
?
?
(xxii)
2
2 2 2 2 1
x a x
a x dx a x sin c
2 2 a
?
? ? ? ? ?
?
(xxiii)
? ?
2
2 2 2 2 2 2
x a
x a dx x a n x x a c
2 2
? ? ? ? ? ? ?
?
?
(xxiv)
? ?
2
2 2 2 2 2 2
x a
x a dx x a n x x a c
2 2
? ? ? ? ? ? ?
?
?
(xxv)
ax
ax
2 2
e
e .sin bx dx (a sin bx b cos bx) c
a b
? ? ?
?
?
 
ax
1
2 2
e b
sin bx tan c
a
a b
?
? ?
? ? ?
? ?
? ? ?
(xxvi)
ax
ax
2 2
e
e .cos bx dx (a cos bx b sin bx) c
a b
? ? ?
?
?
ax
1
2 2
e b
cos bx tan c
a
a b
?
? ?
? ? ?
? ?
? ? ?
3 . TECHNIQUES OF INTEGRATION  :
( a ) Substitution or change of independent variable :
If (x) ? is a continuous differentiable function, then to evaluate integrals of the form f( (x)) '(x)dx ? ?
?
, we
substitute ?(x) = t and ?'(x)dx = dt.
Hence I f( (x)) '(x)dx ? ? ?
?
 reduces to f(t)dt
?
.
(i) Fundamental deductions of method of substitution :
? ?
n
f(x) f '(x)dx
?
OR
n
f '(x)
[f(x)]
? dx   put f(x) = t & proceed.
Illustration 1 : Evaluate 
3
2
cos x
sin x sin x ?
?
dx
Solution : I =
2
(1 sin x) cos x
dx
sin x(1 sin x)
?
?
?
=
1 sin x
cos x dx
sin x
?
?
Put sinx = t  ? cosx dx = dt
?  I =
1 t
dt n| t| t c
t
?
? ? ?
?
? = n| sin x| sin x c ? ? ? Ans.
Illustration 2 : Evaluate 
? ?
? ?
2
4 2 1
x 1 dx
1
x 3x 1 tan x
x
?
?
? ?
? ? ?
? ?
? ?
?
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NODE6\ E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#06\Eng\01-INDEFINITE INTEGRATION.p65
If f & F are function of x such that F' (x) = f(x) then the function F is called a PRIMITIVE OR ANTIDERIVATIVE
OR INTEGRAL of f(x) w.r.t. x and is written symbolically as
d
f(x) dx F(x) c { F(x) c} f(x)
dx
? ? ? ? ?
?
, where c is called the constant of integration.
1 . GEOMETRICAL INTERPRETATION OF INDEFINITE INTEGRAL :
f(x)dx F(x) c y(say) ? ? ?
?
, represents a family of curves. The different values of c will correspond to different
members of this family and these members can be obtained by shifting any one of the curves parallel to itself.
This is the geometrical interpretation of indefinite integral.
Let f(x) = 2x. Then 
2
f(x)dx x c. ? ?
?
 For different values                
Y
X X '
Y '
y=x+3
  
2
y=x+2
  
2
y=x+1
  
2
y=x
  
2
y=x–1
  
2
y=x–2
  
2
y=x–3
  
2
P
3
P
2
P
1
P
0
P
– 1
P
–3
P
–2
x = a
of c, we get different integrals. But these integrals are
very similar geometrically.
Thus, y = x
2 
+ c, where c is arbitrary constant, represents
a family of integrals. By assigning different values to c,
we get different members of the family. These together
constitute the indefinite integral. In this case, each integral
represents a parabola with its axis along y-axis.
If the line x = a intersects the parabolas y = x
2
, y = x
2
 +1,
y = x
2
 + 2, y = x
2
 – 1, y = x
2
 – 2 at P
0
, P
1
, P
2
, P
–1
, P
–2
 etc.,
respectively, then 
dy
dx
 at these points equals 2a. This
indicates that the tangents to the curves at these points
are parallel. Thus, 
?
2
2 xdx = x + c = f(x) + c (say),
implies that the tangents to all the curves
f(x) + c, c ? R, at the points of intersection of the
curves by the line x = a, (a ? R) , are parallel.
2 . STANDARD RESULTS :
(i)
n 1
n
(ax b)
(ax b) dx c; n 1
a(n 1)
?
?
? ? ? ? ?
?
?
(ii)
dx 1
n ax b c
ax b a
? ? ?
?
?
?
(iii)
ax b ax b
1
e dx e c
a
? ?
? ?
?
(iv)
px q
px q
1 a
a dx c, (a 0)
p n a
?
?
? ? ?
?
?
(v)
1
sin(ax b)dx cos(ax b) c
a
? ? ? ? ?
?
(vi)
1
cos(ax b)dx sin(ax b) c
a
? ? ? ?
?
(vii)
1
tan(ax b)dx n| sec(ax b)| c
a
? ? ? ?
?
?
(viii)
1
cot(ax b)dx n| sin(ax b)| c
a
? ? ? ?
?
?
(ix)
2
1
sec (ax b)dx tan(ax b) c
a
? ? ? ?
?
(x)
2
1
cos ec (ax b)dx cot(ax b) c
a
? ? ? ? ?
?
(xi)
1
cosec (ax b).cot(ax b)dx cosec (ax b) c
a
? ? ? ? ? ?
?
(xii) sec
?
(ax + b).tan(ax + b)dx = 
1
sec(ax b) c
a
? ?
INDEFINITE INTEGRATION
JEEMAIN.GURU
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NODE6\ E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#06\Eng\01-INDEFINITE INTEGRATION.p65
(xiii) sec x dx n sec x tan x c ? ? ?
?
?
x
= n tan c
4 2
? ? ?
? ?
? ?
? ?
?
(xiv) cosec x dx n cosec x cot x c ? ? ?
?
? = 
x
n tan c
2
? ? = n | cosec x cot x| ? ? ? + c
(xv)
1
2 2
dx x
sin c
a
a x
?
? ?
?
? (xvi)
1
2 2
dx 1 x
tan c
a a a x
?
? ?
?
?
(xvii)
1
2 2
dx 1 x
sec c
a a
x x a
?
? ?
?
? (xviii)
2 2
2 2
dx
n x x a
x a
? ?
? ? ?
? ?
? ?
?
?
?
 + c
(xix)
2 2
2 2
dx
n x x a
x a
? ?
? ? ?
? ?
? ?
?
?
?
 + c (xx)
2 2
dx 1 a x
n c
2a a x a x
?
? ?
? ?
?
?
(xxi)
2 2
dx 1 x a
n c
2a x a x a
?
? ?
? ?
?
?
(xxii)
2
2 2 2 2 1
x a x
a x dx a x sin c
2 2 a
?
? ? ? ? ?
?
(xxiii)
? ?
2
2 2 2 2 2 2
x a
x a dx x a n x x a c
2 2
? ? ? ? ? ? ?
?
?
(xxiv)
? ?
2
2 2 2 2 2 2
x a
x a dx x a n x x a c
2 2
? ? ? ? ? ? ?
?
?
(xxv)
ax
ax
2 2
e
e .sin bx dx (a sin bx b cos bx) c
a b
? ? ?
?
?
 
ax
1
2 2
e b
sin bx tan c
a
a b
?
? ?
? ? ?
? ?
? ? ?
(xxvi)
ax
ax
2 2
e
e .cos bx dx (a cos bx b sin bx) c
a b
? ? ?
?
?
ax
1
2 2
e b
cos bx tan c
a
a b
?
? ?
? ? ?
? ?
? ? ?
3 . TECHNIQUES OF INTEGRATION  :
( a ) Substitution or change of independent variable :
If (x) ? is a continuous differentiable function, then to evaluate integrals of the form f( (x)) '(x)dx ? ?
?
, we
substitute ?(x) = t and ?'(x)dx = dt.
Hence I f( (x)) '(x)dx ? ? ?
?
 reduces to f(t)dt
?
.
(i) Fundamental deductions of method of substitution :
? ?
n
f(x) f '(x)dx
?
OR
n
f '(x)
[f(x)]
? dx   put f(x) = t & proceed.
Illustration 1 : Evaluate 
3
2
cos x
sin x sin x ?
?
dx
Solution : I =
2
(1 sin x) cos x
dx
sin x(1 sin x)
?
?
?
=
1 sin x
cos x dx
sin x
?
?
Put sinx = t  ? cosx dx = dt
?  I =
1 t
dt n| t| t c
t
?
? ? ?
?
? = n| sin x| sin x c ? ? ? Ans.
Illustration 2 : Evaluate 
? ?
? ?
2
4 2 1
x 1 dx
1
x 3x 1 tan x
x
?
?
? ?
? ? ?
? ?
? ?
?
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Solution : The given integral can be written as
I = 
2
2
1
1
1 dx
x
1 1
x 1 tan x
x x
?
? ?
?
? ?
? ?
? ?
? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ?
? ?
?
Let 
1
x
x
? ?
?
? ?
? ?
= t. Differentiating we get 
2
1
1
x
? ?
?
? ?
? ?
dx = dt
Hence I = 
? ?
2 1
dt
t 1 tan t
?
?
?
Now make one more substitution tan
–1
t = u. Then 
2
dt
t 1 ?
 = du and I = 
du
n| u| c
u
? ?
?
?
Returning to t, and then to x, we have
I = 
1 1
1
n| tan t| c n tan x c
x
? ?
? ?
? ? ? ?
? ?
? ?
? ? Ans.
Do yourself -1 :
(i) Evaluate : 
2
6
x
dx
9 16x ?
?
(ii) Evaluate : 
3
cos x dx
?
(ii) Standard substitutions :
2 2
dx
a x ?
?
 or 
2 2
a x dx ?
?
  ; put  x = a tan ? or x = a cot
?
2 2
dx
a x ?
?
 or 
2 2
a x dx ?
?
  ; put  x = a sin ? or x = a cos
?
2 2
dx
x a ?
? or 
2 2
x a dx ?
?
 ; put  x = a sec ? or x = a cosec
?
a x
dx
a x
?
?
?
 ; put  x = a cos2 ?
x
x
? ?
? ?
?
  dx  or  (x )( x) ? ? ? ?
?
  ; put  x = ? cos
2
 
?
+ ? sin
2
?
x
x
? ?
? ?
?
  dx  or  (x )(x ) ? ? ? ?
?
  ; put x = ? sec
2
?
 – ? tan
2
 
?
dx
(x )(x ) ? ? ? ?
?  ; put  x – ? = t
2
   or  x  – ?  = t
2
.
Illustration 3 : Evaluate 
dx
(x a)(b x) ? ?
?
Solution : Put x = acos
2
? + bsin
2
?, the given integral becomes
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J E E - M a t h e m a t i c s
NODE6\ E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#06\Eng\01-INDEFINITE INTEGRATION.p65
If f & F are function of x such that F' (x) = f(x) then the function F is called a PRIMITIVE OR ANTIDERIVATIVE
OR INTEGRAL of f(x) w.r.t. x and is written symbolically as
d
f(x) dx F(x) c { F(x) c} f(x)
dx
? ? ? ? ?
?
, where c is called the constant of integration.
1 . GEOMETRICAL INTERPRETATION OF INDEFINITE INTEGRAL :
f(x)dx F(x) c y(say) ? ? ?
?
, represents a family of curves. The different values of c will correspond to different
members of this family and these members can be obtained by shifting any one of the curves parallel to itself.
This is the geometrical interpretation of indefinite integral.
Let f(x) = 2x. Then 
2
f(x)dx x c. ? ?
?
 For different values                
Y
X X '
Y '
y=x+3
  
2
y=x+2
  
2
y=x+1
  
2
y=x
  
2
y=x–1
  
2
y=x–2
  
2
y=x–3
  
2
P
3
P
2
P
1
P
0
P
– 1
P
–3
P
–2
x = a
of c, we get different integrals. But these integrals are
very similar geometrically.
Thus, y = x
2 
+ c, where c is arbitrary constant, represents
a family of integrals. By assigning different values to c,
we get different members of the family. These together
constitute the indefinite integral. In this case, each integral
represents a parabola with its axis along y-axis.
If the line x = a intersects the parabolas y = x
2
, y = x
2
 +1,
y = x
2
 + 2, y = x
2
 – 1, y = x
2
 – 2 at P
0
, P
1
, P
2
, P
–1
, P
–2
 etc.,
respectively, then 
dy
dx
 at these points equals 2a. This
indicates that the tangents to the curves at these points
are parallel. Thus, 
?
2
2 xdx = x + c = f(x) + c (say),
implies that the tangents to all the curves
f(x) + c, c ? R, at the points of intersection of the
curves by the line x = a, (a ? R) , are parallel.
2 . STANDARD RESULTS :
(i)
n 1
n
(ax b)
(ax b) dx c; n 1
a(n 1)
?
?
? ? ? ? ?
?
?
(ii)
dx 1
n ax b c
ax b a
? ? ?
?
?
?
(iii)
ax b ax b
1
e dx e c
a
? ?
? ?
?
(iv)
px q
px q
1 a
a dx c, (a 0)
p n a
?
?
? ? ?
?
?
(v)
1
sin(ax b)dx cos(ax b) c
a
? ? ? ? ?
?
(vi)
1
cos(ax b)dx sin(ax b) c
a
? ? ? ?
?
(vii)
1
tan(ax b)dx n| sec(ax b)| c
a
? ? ? ?
?
?
(viii)
1
cot(ax b)dx n| sin(ax b)| c
a
? ? ? ?
?
?
(ix)
2
1
sec (ax b)dx tan(ax b) c
a
? ? ? ?
?
(x)
2
1
cos ec (ax b)dx cot(ax b) c
a
? ? ? ? ?
?
(xi)
1
cosec (ax b).cot(ax b)dx cosec (ax b) c
a
? ? ? ? ? ?
?
(xii) sec
?
(ax + b).tan(ax + b)dx = 
1
sec(ax b) c
a
? ?
INDEFINITE INTEGRATION
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(xiii) sec x dx n sec x tan x c ? ? ?
?
?
x
= n tan c
4 2
? ? ?
? ?
? ?
? ?
?
(xiv) cosec x dx n cosec x cot x c ? ? ?
?
? = 
x
n tan c
2
? ? = n | cosec x cot x| ? ? ? + c
(xv)
1
2 2
dx x
sin c
a
a x
?
? ?
?
? (xvi)
1
2 2
dx 1 x
tan c
a a a x
?
? ?
?
?
(xvii)
1
2 2
dx 1 x
sec c
a a
x x a
?
? ?
?
? (xviii)
2 2
2 2
dx
n x x a
x a
? ?
? ? ?
? ?
? ?
?
?
?
 + c
(xix)
2 2
2 2
dx
n x x a
x a
? ?
? ? ?
? ?
? ?
?
?
?
 + c (xx)
2 2
dx 1 a x
n c
2a a x a x
?
? ?
? ?
?
?
(xxi)
2 2
dx 1 x a
n c
2a x a x a
?
? ?
? ?
?
?
(xxii)
2
2 2 2 2 1
x a x
a x dx a x sin c
2 2 a
?
? ? ? ? ?
?
(xxiii)
? ?
2
2 2 2 2 2 2
x a
x a dx x a n x x a c
2 2
? ? ? ? ? ? ?
?
?
(xxiv)
? ?
2
2 2 2 2 2 2
x a
x a dx x a n x x a c
2 2
? ? ? ? ? ? ?
?
?
(xxv)
ax
ax
2 2
e
e .sin bx dx (a sin bx b cos bx) c
a b
? ? ?
?
?
 
ax
1
2 2
e b
sin bx tan c
a
a b
?
? ?
? ? ?
? ?
? ? ?
(xxvi)
ax
ax
2 2
e
e .cos bx dx (a cos bx b sin bx) c
a b
? ? ?
?
?
ax
1
2 2
e b
cos bx tan c
a
a b
?
? ?
? ? ?
? ?
? ? ?
3 . TECHNIQUES OF INTEGRATION  :
( a ) Substitution or change of independent variable :
If (x) ? is a continuous differentiable function, then to evaluate integrals of the form f( (x)) '(x)dx ? ?
?
, we
substitute ?(x) = t and ?'(x)dx = dt.
Hence I f( (x)) '(x)dx ? ? ?
?
 reduces to f(t)dt
?
.
(i) Fundamental deductions of method of substitution :
? ?
n
f(x) f '(x)dx
?
OR
n
f '(x)
[f(x)]
? dx   put f(x) = t & proceed.
Illustration 1 : Evaluate 
3
2
cos x
sin x sin x ?
?
dx
Solution : I =
2
(1 sin x) cos x
dx
sin x(1 sin x)
?
?
?
=
1 sin x
cos x dx
sin x
?
?
Put sinx = t  ? cosx dx = dt
?  I =
1 t
dt n| t| t c
t
?
? ? ?
?
? = n| sin x| sin x c ? ? ? Ans.
Illustration 2 : Evaluate 
? ?
? ?
2
4 2 1
x 1 dx
1
x 3x 1 tan x
x
?
?
? ?
? ? ?
? ?
? ?
?
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Solution : The given integral can be written as
I = 
2
2
1
1
1 dx
x
1 1
x 1 tan x
x x
?
? ?
?
? ?
? ?
? ?
? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ?
? ?
?
Let 
1
x
x
? ?
?
? ?
? ?
= t. Differentiating we get 
2
1
1
x
? ?
?
? ?
? ?
dx = dt
Hence I = 
? ?
2 1
dt
t 1 tan t
?
?
?
Now make one more substitution tan
–1
t = u. Then 
2
dt
t 1 ?
 = du and I = 
du
n| u| c
u
? ?
?
?
Returning to t, and then to x, we have
I = 
1 1
1
n| tan t| c n tan x c
x
? ?
? ?
? ? ? ?
? ?
? ?
? ? Ans.
Do yourself -1 :
(i) Evaluate : 
2
6
x
dx
9 16x ?
?
(ii) Evaluate : 
3
cos x dx
?
(ii) Standard substitutions :
2 2
dx
a x ?
?
 or 
2 2
a x dx ?
?
  ; put  x = a tan ? or x = a cot
?
2 2
dx
a x ?
?
 or 
2 2
a x dx ?
?
  ; put  x = a sin ? or x = a cos
?
2 2
dx
x a ?
? or 
2 2
x a dx ?
?
 ; put  x = a sec ? or x = a cosec
?
a x
dx
a x
?
?
?
 ; put  x = a cos2 ?
x
x
? ?
? ?
?
  dx  or  (x )( x) ? ? ? ?
?
  ; put  x = ? cos
2
 
?
+ ? sin
2
?
x
x
? ?
? ?
?
  dx  or  (x )(x ) ? ? ? ?
?
  ; put x = ? sec
2
?
 – ? tan
2
 
?
dx
(x )(x ) ? ? ? ?
?  ; put  x – ? = t
2
   or  x  – ?  = t
2
.
Illustration 3 : Evaluate 
dx
(x a)(b x) ? ?
?
Solution : Put x = acos
2
? + bsin
2
?, the given integral becomes
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? ?
1
2 2 2 2
2
2(b a) sin cos d
I
(a cos b sin a)(b a cos b sin
? ? ? ?
?
? ? ? ? ? ? ? ?
?
  =
? ?
2(b a) sin cos d
b a sin cos
? ? ? ?
? ? ?
?
= 
1
b a x a
2d 2 c 2 sin c
b a b a
?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ?
?
Ans.
Illustration 4 : Evaluate 
1 x 1
. dx
x
1 x
?
?
?
Solution : Put x = cos
2
?  ? dx = –2sin ? cos ? d ?
?
2
1 cos 1
I . ( 2 sin cos )d 2 tan tan d
1 cos 2
cos
? ? ?
? ? ? ? ? ? ? ? ?
? ?
?
? ?
 
2
sin ( / 2) 1 cos
4 d 2 d 2 n| sec tan | 2 c
cos cos
? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ?
?
 
1
1 1 x
2 n 2 cos x c
x
?
? ?
? ? ? ? ?
Do yourself -2 :
(i) Evaluate : 
x 3
dx
2 x
?
?
?
(ii) Evaluate : 
2
dx
x x 4 ?
?
(b ) Integration by part : 
du
u.v dx u v dx . v dx
dx
? ?
? ?
? ?
? ?
? ? ? ?
 dx where u & v are differentiable functions and
are commonly designated as first & second function respectively.
Note :  While using integration  by parts, choose u & v such that
(i) vdx
?
& (ii)   
du
. v dx
dx
? ?
? ?
? ?
? ?
 dx  are simple to integrate.
This is generally obtained by choosing first function as the function which comes first in the word ILATE,
where; I-Inverse function, L-Logarithmic function, A-Algebraic function, T-Trigonometric function &
E-Exponential function.
Illustration 5 : Evaluate : cos x dx
?
Solution : Consider I = cos xdx
?
Let
x t ?
then
1
dx dt
2 x
?
i.e.
dx 2 xdt ?
or dx = 2t dt
so I cos t.2tdt ?
?
taking t as first function, then integrate it by part
? I =
dt
2 t cos tdt cos tdt dt
dt
? ? ? ?
?
? ?
? ?
? ? ? ?
? ? ?
 = 2 t sin t 1.sin tdt
? ?
?
? ?
?
 = 
? ?
2 t sin t cos t c ? ?
 I = 2 x sin x cos x c
? ?
? ? ? ?
Ans.
Illustration 6 : Evaluate : 
x
dx
1 sin x ?
?
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If f & F are function of x such that F' (x) = f(x) then the function F is called a PRIMITIVE OR ANTIDERIVATIVE
OR INTEGRAL of f(x) w.r.t. x and is written symbolically as
d
f(x) dx F(x) c { F(x) c} f(x)
dx
? ? ? ? ?
?
, where c is called the constant of integration.
1 . GEOMETRICAL INTERPRETATION OF INDEFINITE INTEGRAL :
f(x)dx F(x) c y(say) ? ? ?
?
, represents a family of curves. The different values of c will correspond to different
members of this family and these members can be obtained by shifting any one of the curves parallel to itself.
This is the geometrical interpretation of indefinite integral.
Let f(x) = 2x. Then 
2
f(x)dx x c. ? ?
?
 For different values                
Y
X X '
Y '
y=x+3
  
2
y=x+2
  
2
y=x+1
  
2
y=x
  
2
y=x–1
  
2
y=x–2
  
2
y=x–3
  
2
P
3
P
2
P
1
P
0
P
– 1
P
–3
P
–2
x = a
of c, we get different integrals. But these integrals are
very similar geometrically.
Thus, y = x
2 
+ c, where c is arbitrary constant, represents
a family of integrals. By assigning different values to c,
we get different members of the family. These together
constitute the indefinite integral. In this case, each integral
represents a parabola with its axis along y-axis.
If the line x = a intersects the parabolas y = x
2
, y = x
2
 +1,
y = x
2
 + 2, y = x
2
 – 1, y = x
2
 – 2 at P
0
, P
1
, P
2
, P
–1
, P
–2
 etc.,
respectively, then 
dy
dx
 at these points equals 2a. This
indicates that the tangents to the curves at these points
are parallel. Thus, 
?
2
2 xdx = x + c = f(x) + c (say),
implies that the tangents to all the curves
f(x) + c, c ? R, at the points of intersection of the
curves by the line x = a, (a ? R) , are parallel.
2 . STANDARD RESULTS :
(i)
n 1
n
(ax b)
(ax b) dx c; n 1
a(n 1)
?
?
? ? ? ? ?
?
?
(ii)
dx 1
n ax b c
ax b a
? ? ?
?
?
?
(iii)
ax b ax b
1
e dx e c
a
? ?
? ?
?
(iv)
px q
px q
1 a
a dx c, (a 0)
p n a
?
?
? ? ?
?
?
(v)
1
sin(ax b)dx cos(ax b) c
a
? ? ? ? ?
?
(vi)
1
cos(ax b)dx sin(ax b) c
a
? ? ? ?
?
(vii)
1
tan(ax b)dx n| sec(ax b)| c
a
? ? ? ?
?
?
(viii)
1
cot(ax b)dx n| sin(ax b)| c
a
? ? ? ?
?
?
(ix)
2
1
sec (ax b)dx tan(ax b) c
a
? ? ? ?
?
(x)
2
1
cos ec (ax b)dx cot(ax b) c
a
? ? ? ? ?
?
(xi)
1
cosec (ax b).cot(ax b)dx cosec (ax b) c
a
? ? ? ? ? ?
?
(xii) sec
?
(ax + b).tan(ax + b)dx = 
1
sec(ax b) c
a
? ?
INDEFINITE INTEGRATION
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(xiii) sec x dx n sec x tan x c ? ? ?
?
?
x
= n tan c
4 2
? ? ?
? ?
? ?
? ?
?
(xiv) cosec x dx n cosec x cot x c ? ? ?
?
? = 
x
n tan c
2
? ? = n | cosec x cot x| ? ? ? + c
(xv)
1
2 2
dx x
sin c
a
a x
?
? ?
?
? (xvi)
1
2 2
dx 1 x
tan c
a a a x
?
? ?
?
?
(xvii)
1
2 2
dx 1 x
sec c
a a
x x a
?
? ?
?
? (xviii)
2 2
2 2
dx
n x x a
x a
? ?
? ? ?
? ?
? ?
?
?
?
 + c
(xix)
2 2
2 2
dx
n x x a
x a
? ?
? ? ?
? ?
? ?
?
?
?
 + c (xx)
2 2
dx 1 a x
n c
2a a x a x
?
? ?
? ?
?
?
(xxi)
2 2
dx 1 x a
n c
2a x a x a
?
? ?
? ?
?
?
(xxii)
2
2 2 2 2 1
x a x
a x dx a x sin c
2 2 a
?
? ? ? ? ?
?
(xxiii)
? ?
2
2 2 2 2 2 2
x a
x a dx x a n x x a c
2 2
? ? ? ? ? ? ?
?
?
(xxiv)
? ?
2
2 2 2 2 2 2
x a
x a dx x a n x x a c
2 2
? ? ? ? ? ? ?
?
?
(xxv)
ax
ax
2 2
e
e .sin bx dx (a sin bx b cos bx) c
a b
? ? ?
?
?
 
ax
1
2 2
e b
sin bx tan c
a
a b
?
? ?
? ? ?
? ?
? ? ?
(xxvi)
ax
ax
2 2
e
e .cos bx dx (a cos bx b sin bx) c
a b
? ? ?
?
?
ax
1
2 2
e b
cos bx tan c
a
a b
?
? ?
? ? ?
? ?
? ? ?
3 . TECHNIQUES OF INTEGRATION  :
( a ) Substitution or change of independent variable :
If (x) ? is a continuous differentiable function, then to evaluate integrals of the form f( (x)) '(x)dx ? ?
?
, we
substitute ?(x) = t and ?'(x)dx = dt.
Hence I f( (x)) '(x)dx ? ? ?
?
 reduces to f(t)dt
?
.
(i) Fundamental deductions of method of substitution :
? ?
n
f(x) f '(x)dx
?
OR
n
f '(x)
[f(x)]
? dx   put f(x) = t & proceed.
Illustration 1 : Evaluate 
3
2
cos x
sin x sin x ?
?
dx
Solution : I =
2
(1 sin x) cos x
dx
sin x(1 sin x)
?
?
?
=
1 sin x
cos x dx
sin x
?
?
Put sinx = t  ? cosx dx = dt
?  I =
1 t
dt n| t| t c
t
?
? ? ?
?
? = n| sin x| sin x c ? ? ? Ans.
Illustration 2 : Evaluate 
? ?
? ?
2
4 2 1
x 1 dx
1
x 3x 1 tan x
x
?
?
? ?
? ? ?
? ?
? ?
?
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Solution : The given integral can be written as
I = 
2
2
1
1
1 dx
x
1 1
x 1 tan x
x x
?
? ?
?
? ?
? ?
? ?
? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ?
? ?
?
Let 
1
x
x
? ?
?
? ?
? ?
= t. Differentiating we get 
2
1
1
x
? ?
?
? ?
? ?
dx = dt
Hence I = 
? ?
2 1
dt
t 1 tan t
?
?
?
Now make one more substitution tan
–1
t = u. Then 
2
dt
t 1 ?
 = du and I = 
du
n| u| c
u
? ?
?
?
Returning to t, and then to x, we have
I = 
1 1
1
n| tan t| c n tan x c
x
? ?
? ?
? ? ? ?
? ?
? ?
? ? Ans.
Do yourself -1 :
(i) Evaluate : 
2
6
x
dx
9 16x ?
?
(ii) Evaluate : 
3
cos x dx
?
(ii) Standard substitutions :
2 2
dx
a x ?
?
 or 
2 2
a x dx ?
?
  ; put  x = a tan ? or x = a cot
?
2 2
dx
a x ?
?
 or 
2 2
a x dx ?
?
  ; put  x = a sin ? or x = a cos
?
2 2
dx
x a ?
? or 
2 2
x a dx ?
?
 ; put  x = a sec ? or x = a cosec
?
a x
dx
a x
?
?
?
 ; put  x = a cos2 ?
x
x
? ?
? ?
?
  dx  or  (x )( x) ? ? ? ?
?
  ; put  x = ? cos
2
 
?
+ ? sin
2
?
x
x
? ?
? ?
?
  dx  or  (x )(x ) ? ? ? ?
?
  ; put x = ? sec
2
?
 – ? tan
2
 
?
dx
(x )(x ) ? ? ? ?
?  ; put  x – ? = t
2
   or  x  – ?  = t
2
.
Illustration 3 : Evaluate 
dx
(x a)(b x) ? ?
?
Solution : Put x = acos
2
? + bsin
2
?, the given integral becomes
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? ?
1
2 2 2 2
2
2(b a) sin cos d
I
(a cos b sin a)(b a cos b sin
? ? ? ?
?
? ? ? ? ? ? ? ?
?
  =
? ?
2(b a) sin cos d
b a sin cos
? ? ? ?
? ? ?
?
= 
1
b a x a
2d 2 c 2 sin c
b a b a
?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ?
?
Ans.
Illustration 4 : Evaluate 
1 x 1
. dx
x
1 x
?
?
?
Solution : Put x = cos
2
?  ? dx = –2sin ? cos ? d ?
?
2
1 cos 1
I . ( 2 sin cos )d 2 tan tan d
1 cos 2
cos
? ? ?
? ? ? ? ? ? ? ? ?
? ?
?
? ?
 
2
sin ( / 2) 1 cos
4 d 2 d 2 n| sec tan | 2 c
cos cos
? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ?
?
 
1
1 1 x
2 n 2 cos x c
x
?
? ?
? ? ? ? ?
Do yourself -2 :
(i) Evaluate : 
x 3
dx
2 x
?
?
?
(ii) Evaluate : 
2
dx
x x 4 ?
?
(b ) Integration by part : 
du
u.v dx u v dx . v dx
dx
? ?
? ?
? ?
? ?
? ? ? ?
 dx where u & v are differentiable functions and
are commonly designated as first & second function respectively.
Note :  While using integration  by parts, choose u & v such that
(i) vdx
?
& (ii)   
du
. v dx
dx
? ?
? ?
? ?
? ?
 dx  are simple to integrate.
This is generally obtained by choosing first function as the function which comes first in the word ILATE,
where; I-Inverse function, L-Logarithmic function, A-Algebraic function, T-Trigonometric function &
E-Exponential function.
Illustration 5 : Evaluate : cos x dx
?
Solution : Consider I = cos xdx
?
Let
x t ?
then
1
dx dt
2 x
?
i.e.
dx 2 xdt ?
or dx = 2t dt
so I cos t.2tdt ?
?
taking t as first function, then integrate it by part
? I =
dt
2 t cos tdt cos tdt dt
dt
? ? ? ?
?
? ?
? ?
? ? ? ?
? ? ?
 = 2 t sin t 1.sin tdt
? ?
?
? ?
?
 = 
? ?
2 t sin t cos t c ? ?
 I = 2 x sin x cos x c
? ?
? ? ? ?
Ans.
Illustration 6 : Evaluate : 
x
dx
1 sin x ?
?
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Solution : Let I =
x
dx
1 sin x ?
?
 = 
x(1 sin x)
dx
(1 sin x)(1 sin x)
?
? ?
?
  = 
2
x(1 sin x)
dx
1 sin x
?
?
?
 = 
2
2
x(1 sin x)
dx x sec xdx x sec x tan xdx
cos x
?
? ?
? ? ?
  = 
2 2
dx
x sec xdx sec xdx dx
dx
? ? ? ?
?
? ?
? ?
? ? ? ?
? ? ? – 
dx
x sec x tan xdx sec x tan xdx dx
dx
? ? ? ?
?
? ?
? ?
? ? ? ?
? ? ?
  = x tan x tan xdx x sec x sec xdx
? ? ? ?
? ? ?
? ? ? ?
? ?
  = ? ? ? ?
x tan x n| sec x| x sec x n| sec x tan x| c ? ? ? ? ? ? ?
  = x
? ?
(sec x tan x)
tan x sec x n c
sec x
?
? ? ? ?
=
x(1 sin x)
n|1 sin x| c
cos x
? ?
? ? ? ? Ans.
Do yourself -3 :
(i) Evaluate : 
x
xe dx
?
(ii) Evaluate : 
3 2
x sin(x )dx
?
Two classic integrands :
(i)
x x
e [f(x) f '(x)]dx e .f(x) c ? ? ?
?
Illustration 7 : Evaluate 
2
x
2
1 x
e dx
1 x
? ? ?
? ?
? ?
?
?
Solution :
2
x
2
1 x
e dx
1 x
? ? ?
? ?
? ?
?
?
=
2
x
2 2
(1 2x x )
e dx
(1 x )
? ?
?
?
=
x
x
2 2 2 2
1 2x e
e dx c
(1 x ) (1 x ) 1 x
? ?
? ? ?
? ?
? ? ?
? ?
? Ans.
Illustration 8 : The value of 
4
x
2 5 / 2
x 2
e dx
(1 x )
? ? ?
? ?
? ? ?
?
 is equal to -
(A) 
x
2 3 / 2
e (x 1)
(1 x )
?
?
(B) 
x 2
2 3 / 2
e (1 x x )
(1 x )
? ?
?
(C) 
x
2 3 / 2
e (1 x)
(1 x )
?
?
(D) none of these
Solution : Let I = 
4
x
2 5 / 2
x 2
e dx
(1 x )
? ? ?
? ?
? ? ?
?
 = 
2
x
2 1 / 2 2 5 / 2
1 1 2x
e .dx
(1 x ) (1 x )
? ? ?
?
? ?
? ? ? ?
?
= 
2
x
2 1 / 2 2 3 / 2 2 3 / 2 2 5 / 2
1 x x 1 2x
e dx
(1 x ) (1 x ) (1 x ) (1 x )
? ? ?
? ? ?
? ?
? ? ? ? ? ?
?
= 
x x
2 1 / 2 2 3 / 2
e xe
c
(1 x ) (1 x )
? ?
? ?
=
x 2
2 3 / 2
e { 1 x x}
c
(1 x )
? ?
?
?
Ans. (D)
Do yourself -4 :
(i) Evaluate : 
x 1
2
1
e tan x dx
1 x
?
? ?
?
? ?
?
? ?
?
(ii) Evaluate : 
? ?
2
x 2 2
xe sin x cos x dx ?
?
(ii) [f(x) xf '(x)]dx x f(x) c ? ? ?
?
Illustration
 
 9
 
: Evaluate 
x sin x
dx
1 cos x
?
?
?
JEEMAIN.GURU
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