MATHS
Indefinite Integration If f & g are functions of x such that g(x) = f(x), then indefinite integration of f(x) with respect to x is defined and denoted as
f(x) dx = g(x) + C, where C is called the constant of integration.
Standard Formula:
ax b n 1 dx = a n 1
(i)
(ax + b)n
(ii)
dx 1 = n |ax + b| + C ax b a
(iii)
eax+b dx =
1 ax+b e +C a
(iv)
apx+q dx =
1 a pxq + C; a > 0 p n a
(v)
sin (ax + b) dx =
(vi)
cos (ax + b) dx =
1 sin (ax + b) + C a
(vii)
tan(ax + b) dx =
1 n |sec (ax + b)| + C a
cot(ax + b) dx =
1 n |sin(ax + b)| + C a
(viii)
+ C, n
1
1 cos (ax + b) + C a
1 tan(ax + b) + C a
(ix)
sec² (ax + b) dx =
(x)
cosec²(ax + b) dx = –
(xi)
sec (ax + b). tan (ax + b) dx =
(xii)
cosec (ax + b). cot (ax + b) dx = –
(xiii)
secx dx = n |secx + tanx| + C
(xiv)
cosec x dx = n |cosecx
1 cot(ax + b)+ C a 1 sec (ax + b) + C a 1 cosec (ax + b) + C a x n tan + C 4 2
OR
cotx| + C OR n tan
x + C OR 2
n |cosecx + cotx| + C
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MATHS (xv)
(xvi)
(xvii)
(xviii)
(xix)
(xx)
ax dx 1 ln ax + C 2 = a x 2a
(xxi)
xa dx 1 ln xa + C 2 = x a 2a
(xxii)
(xxiv)
(xxvi)
a x 2
= sin1
2
1 x dx = tan1 +C 2 a a a x
dx x x a 2
dx x2 a2 dx x a 2
2
1 x sec 1 +C a a
=
2
2 2 = n x x a + C
2 2 = n x x a + C
OR
OR
sinh1
x +C a
cosh1
x +C a
2
2
a 2 x 2 dx =
x 2
a 2 x2 +
x 2
x 2 a 2 dx =
x x a dx = 2 2
2
eax. sin bx dx =
x +C a
2
(xxiii)
(xxv)
dx
a2 x sin1 +C 2 a
x2 a 2 +
x a 2
2
a2 x x 2 a2 n +C a 2
x x 2 a2 a2 n +C a 2
e ax (a sin bx b cos bx) + C a 2 b2
eax. cos bx dx =
e ax (a cos bx + b sin bx) + C a 2 b2
Theorems on integration (i) (ii) (iii)
C f (x).dx = C f (x).dx (f (x) g(x)) dx = f ( x)dx g(x) dx f (x)dx g( x) C
1
f (ax b)dx =
g(ax b) + C2 a
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MATHS Example # 1
Evaluate :
Solution.
4x
Example # 2
Evaluate :
Solution.
x
5
4x
3
=
x
=
x
3
3
dx
4 6 2 6 x +C= x + C. 6 3
dx =
5
x
5x 2 4
3
7 2 dx x x
5x 2 4
5x
dx +
2
7 dx + x
dx – 4 . 1 . dx + 7 .
4dx +
dx –
x
dx + 5 .
7 2 dx x x
2
2 x
dx
1 dx + 2 . x
x
1/ 2
dx
x 1/ 2 x3 x4 = +5. – 4x + 7 n | x | + 2 1/ 2 + C 3 4 =
5 3 x4 + x – 4x + 7 n | x | + 4 3 4
Example # 3
Evaluate :
Solution.
We have,
(e
xn a
e
xn a
e anx eana dx , a > 0
e anx e ana ) dx =
=
Example # 4
Evaluate : 2x 3x
Solution.
Example # 5
Evaluate :
Solution.
5x
2x 3x
dx =
5x
sin
3
+C
x
(e
na x
a
a
e nx e na ) dx =
a x dx +
x a dx +
(a
a a dx =
x
x a a a ) dx
ax x a 1 + + aa . x + C. n a a 1
dx
2x 3x 5x 5x
dx =
x x 2 3 (2 / 5) x (3 / 5 ) x dx = + +C 2 5 5 3 n n 5 5
x cos3 x dx
sin3 x cos3 x dx = =
1 8 1 32
(2 sin x cos x)
3
dx =
1 8
(3 sin 2x sin 6x) dx =
sin3 2x dx =
1 8
3 sin 2x sin 6 x dx 4
1 1 3 2 cos 2x 6 cos 6 x + C 32
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MATHS Example # 6
Solution.
Evaluate :
x
x4 2
1
dx
2
1
dx =
x4 1 1
=
Example # 7
Evaluate :
Solution.
We have
1
4 9x
x
x4
2
x2 1
dx =
( x 2 1) dx +
1
4 9x
2
1 9
dx =
x4 1 1 dx x 2 1 x 2 1
x
1 2
1
dx =
x3 – x + tan–1 x + C 3
dx
1
4x
dx = 2
1 9
1
( 2 / 3)
2
x2
dx
9
1 x 3x 1 1 +C= +C . tan–1 tan–1 ( 2 / 3) 9 6 2/3 2
=
cos x cos 2x dx
Example # 8
Evaluate :
Solution.
cos x cos 2x dx = 2 2 cos x cos 2x dx 1
=
1 2
(cos 3x cos x) dx
=
1 sin 3 x sin x +C 2 3
Self Practice Problems (1)
Evaluate :
tan
(2)
Evaluate :
1 sin x
2
x dx
1
Answers : (1) tanx – x + C
dx (2) tanx – sec x + C
Integration by Substitution If we substitution (x) = t in an integral then (i) everywhere x will be replaced in terms of new variable t. (ii) dx also gets converted in terms of dt. Example # 9
Evaluate :
Solution.
We have =
x
3
1 4
3
sin x 4 dx
sin x 4 dx
Let x 4 = t =
x
sin t dt = –
d(x 4) = dt
4x 3 dx = dt dx =
1 4x 3
dt
1 1 cost + C = – cos x 4 + C 4 4
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MATHS
Example # 10 Evaluate : Solution.
(n x )2 dx x
Let =
=
Let =
(1 sin
=
Solution.
x ) cos x dx
Put sinx = t
x
x
x
dx =
x 1
4
2
1 2
=
1 = . 2
t3 sin3 x + c = sin x + +C 3 3
dx
(x
1 t2 t 1
1 3 2
tan
–1
x ) x2 1
2 2
dt
=
1 2
dx
2 3 1 t 2 2
t 1 1 2 +C = tan–1 3 3 2
2
(iii)
dx ; n N Take x n common & put 1 + x n = t. n x ( x 1)
(iv)
(v)
x x 1
( n1)
dx
x 1 x n
; n
2x 2 1 + C. 3
( f ( x ))n1 +C n 1
f (x) ( f ( x ))1n dx = +C,n1 1 n f (x)n
dx
dt } 2
dt
2t 1 1 +C= tan–1 3 3
(ii)
n
x.dx =
1
2
{Put x 2 = t
[ f(x)]n f (x) dx =
(i)
cosx dx = dt
We have, =
Note:
x ) cos x dx
x2 1
4
1 dx = dt x
t3 ( n x )3 +c = +C 3 3
(1 t 2 ) dt = t +
x
Example # 12 Evaluate :
2
2
Put nx = t
t 2dt =
(1 sin
Example # 11 Evaluate : Solution.
(n x )2 dx x
N, take xn common & put 1 + xn = tn
n
n 1/ n
; take x n common as x and put 1 + x n = t.
Self Practice Problems (3)
Evaluate :
sec 2 x dx 1 tan x
(4)
Evaluate :
sin(nx ) dx x
Answers : (3) n |1 + tan x| + C
(4) – cos (n x) + C
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MATHS Integration by Parts : Product of two functions f(x) and g(x) can be integrate using formula :
f( x) g(x) dx
= f(x)
d
g(x) dx – dx f ( x) g( x) dx dx g( x) dx then it will not contain arbitarary constant.
(i)
when you find integral
(ii)
g( x) dx should be taken as same at both places.
(iii)
The choice of f(x) and g(x) can be decided by ILATE guideline. the function will come later is taken an integral function (g(x)). Inverse function L Logarithmic function A Algebraic function T Trigonometric function E Exponential function
Example # 13 Evaluate : Solution.
Let =
x tan
= (tan–1 x)
1
1
x dx
x dx
x2 – 2
1
1 x
2
.
x2 dx 2
x2 1 1
=
1 x2 tan–1 x – 2 2
=
1 x2 tan–1 x – [x – tan–1 x] + C. 2 2
Example # 14 Evaluate : Solution.
x tan
Let =
x2 1
dx =
1 x2 tan–1 x – 2 2
1 x
dx 1
1 2
x n(1 x) dx
x n(1 x) dx
= n (x + 1) .
x2 – 2
=
1 x2 n (x + 1) – 2 2
=
1 x2 . dx x 1 2
1 x2 x2 dx = n (x + 1) – 2 x 1 2
1 x2 n (x + 1) – 2 2
x2 1 1 x 1 x 1 dx
=
1 x2 n (x + 1) – 2 2
( x 1) x 1 dx
=
2 1 x x n | x 1 | x2 +C n (x + 1) – 2 2 2
x 2 1 1 dx x 1
1
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MATHS
e
Example # 15 Evaluate : Solution.
Let =
e
2x
2x
sin 3x dx
sin 3x dx
cos 3 x – = e2x 3
2e
cos 3 x dx 3
2x
=–
1 2x 2 e cos 3x + 3 3
=–
sin 3 x 1 2x 2 2 x sin 3 x 2e 2 x dx e cos 3x + e 3 3 3 3
=–
1 2x 2 4 e cos 3x + e2x sin 3x – 3 9 9
e
2x
cos 3x dx
e
2x
sin 3x dx
1 2x 2 4 e cos 3x + e2x sin 3x – 3 9 9
=–
+
13 e 2x = (2 sin 3x – 3 cos 3x) 9 9
=
4 e 2x = (2 sin 3x – 3 cos 3x) 9 9
e 2x (2 sin 3x – 3 cos 3x) + C 13
Note : (i) (ii)
ex [f(x) + f (x)] dx = ex. f(x) + C [f(x) + xf (x)] dx = x f(x) + C
Example # 16 Evaluate :
Solution.
x
x
( x 1)2
Given integral =
Example # 17 Evaluate :
Solution.
e
e
Given integral
x
dx
x 1 1
ex
( x 1)2
dx =
1 ex 1 dx = ex +C ( x 1) ( x 1)2 ( x 1)
1 sin x dx 1 cos x
=
e
=
e
x
x
x x 1 2 sin cos 2 2 dx 2 x 2 sin 2 x x x 1 cos ec 2 cot dx = – ex cot +C 2 2 2 2
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MATHS Example # 18 Evaluate :
Solution.
1
n (nx ) (nx)
1
Let =
n (nx ) (nx)
=
e
t
2
2
dx
dx {put x = et et dt}
1 nt 2 dt = t
e
t
1 1 1 nt 2 dt t t t
1 1 = et nt + C = x n (nx ) +C t nx
Self Practice Problems
(5)
Evaluate :
x sin x dx
(6)
Evaluate :
x e
2 x
dx
Answers : (5) – x cosx + sin x + C
(6) x 2 ex – 2xex + 2ex + C
Integration of Rational Algebraic Functions by using Partial Fractions: PARTIAL FRACTIONS :
f (x) If f(x) and g(x) are two polynomials, then g( x ) defines a rational algebraic function of x. f (x) If degree of f(x) < degree of g(x), then g( x ) is called a proper rational function. f (x) If degree of f(x) degree of g(x) then g( x ) is called an improper rational function. f (x) f (x) If g( x ) is an improper rational function, we divide f(x) by g(x) so that the rational function g( x ) is ( x ) expressed in the form (x) + g( x ) , where (x) and ( x ) are polynomials such that the degree of ( x ) f (x) is less than that of g(x). Thus, g( x ) is expressible as the sum of a polynomial and a proper rational function.
f (x) Any proper rational function g( x ) can be expressed as the sum of rational functions, each having a simple factor of g(x). Each such fraction is called a partial fraction and the process of obtained them is
f (x) called the resolutions or decomposition of g( x ) into partial fractions. f (x) The resolution of g( x ) into partial fractions depends mainly upon the nature of the factors of g(x) as discussed below : CASE I When denominator is expressible as the product of non-repeating linear factors. Let g(x) = (x – a1) (x – a2) .....(x – an). Then, we assume that
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MATHS A1 A2 An f (x) + + ..... + g( x ) x a1 x a2 x an where A1, A2, ...... An are constants and can be determined by equating the numerator on R.H.S. to the numerator on L.H.S. and then substituting x = a1, a2, ........,an. Example # 19 Resolve
Solution.
3x 2 x 6 x 2 11x 6 3
into partial fractions.
3x 2
We have,
x 3 6 x 2 11x 6
=
3x 2 ( x 1)( x 2)( x 3)
Let
3x 2 C A B = + + . Then, ( x 1)( x 2)( x 3) x 3 x 1 x2
3x 2 A( x 2)( x 3) B( x 1)( x 3) C( x 1)( x 2) = ( x 1)( x 2)( x 3) ( x 1)( x 2)( x 3)
3x + 2 = A(x – 2) (x – 3) + B (x – 1) (x – 3) + C(x – 1) (x – 2) Putting x – 1 = 0 or x = 1 in (i), we get 5 = A(1 – 2) (1 – 3) A =
...........(i)
5 , 2
Putting x – 2 = 0 or, x = 2 in (i), we obtain 8 = B (2 – 1) (2 – 3) B = –8. Putting x – 3 = 0 or, x = 3 in (i), we obtain 11 = C (3 – 1) (3 – 2) C =
3x 2
x 6 x 11x 6 3
2
=
11 . 2
3x 2 5 11 8 = – + ( x 1)( x 2)( x 3) 2( x 1) 2 ( x 3) x2
Note : In order to determine the value of constants in the numerator of the partial fraction corresponding to the non-repeated linear factor (px + q) in the denominator of a rational expression, we may proceed as follows : Replace x by –
q (obtained by putting px + q = 0) everywhere in the given rational expression except p
in the factor px + q itself. For example, in the above illustration the value of A is obtained by replacing x by 1 in all factors of
A=
3x 2 except (x – 1) i.e. ( x 1)( x 2)( x 3)
3 1 2 5 = (1 2)(1 3) 2
Similarly, we have B=
3 2 1 33 2 11 = –8 and, C = = (1 2)(2 3) (3 1)(3 2) 2
Example # 20 Resolve Solution.
x 3 6 x 2 10 x 2 x 2 5x 6
into partial fractions.
Here the given function is an improper rational function. On dividing we get
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MATHS x 3 6 x 2 10 x 2 x 5x 6 2
we have,
So, let
( x 4) =x–1+
x 4 x 5x 6 2
...........(i)
( x 5 x 6) 2
x 4 ( x 2)( x 3)
=
x 4 B A = + , then ( x 2)( x 3) x 3 x2
– x + 4 = A(x – 3) + B(x – 2) ...........(ii) Putting x – 3 = 0 or x = 3 in (ii), we get 1 = B(1) B = 1. Putting x – 2 = 0 or x = 2 in (ii), we get 2 = A (2 – 3) A = – 2
x 4 1 2 = + ( x 2)( x 3) x 3 x2
x 3 6 x 2 10 x 2
Hence
=x–1–
x 5x 6 2
1 2 + x 3 x2
CASE II When the denominator g(x) is expressible as the product of the linear factors such that some of them are repeating. 1 1 Example g( x ) = this can be expressed as k ( x a) ( x a1 )( x a 2 ).......( x a r )
A3 A2 Ak B1 B2 Br A1 + 2 + 3 + ....+ k + ( x a ) + ( x a ) + ...... + ( x a ) ( x a) ( x a) ( x a) x a 1 2 r Now to determine constants we equate numerators on both sides. Some of the constants are determined by substitution as in case I and remaining are obtained by equating the coefficient of same power of x. The following example illustrate the procedure.
3x 2 Example # 21 Resolve
(3 x 2)dx
( x 1) ( x 1)( x 2) 2
3x 2 Solution.
Let
( x 1) ( x 1)( x 2) 2
=
into partial fractions, and evaluate
( x 1) (x 1)(x 2) 2
A2 A3 A1 A4 + + 2 + ( x 1) x 1 x 1 x2
3x – 2 = A1 (x – 1) (x + 1) (x + 2) + A2 (x + 1) (x + 2) + A3 (x – 1)2 (x + 2) + A4 (x – 1)2 (x + 1) .......(i) Putting x – 1 = 0 or, x = 1 in (i) we get 1 1 = A2 (1 + 1) (1 + 2) A2 = 6 Putting x + 1 = 0 or, x = –1 in (i) we get 5 – 5 = A3 (–2)2 (–1 + 2) A3 = – 4 Putting x + 2 = 0 or, x = –2 in (i) we get 8 – 8 = A4 (–3)2 (–1) A4 = 9 Now equating coefficient of x 3 on both sides, we get 0 = A1 + A3 + A4
A1 = –A3 – A4 =
8 13 5 – = 9 36 4
3x 2 ( x 1) ( x 1)( x 2)
and hence
2
=
1 13 5 8 + + 2 – 36( x 1) 4( x 1) 9( x 2) 6( x 1)
(3 x 2)dx
( x 1) (x 1)(x 2) 2
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MATHS =
1 13 8 5 n |x – 1| – – n |x + 1| + n |x + 2| + C 6 ( x 1 ) 36 9 4
CASE III When some of the factors of denominator g(x) are quadratic but non-repeating. Corresponding
Ax B
to each quadratic factor ax 2 + bx + c, we assume partial fraction of the type
, where A and ax bx c B are constants to be determined by comparing coefficients of similar powers of x in the numerator of both sides. In practice it is advisable to assume partial fractions of the type
2
A(2ax b) ax 2 bx c
+
B ax 2 bx c
The following example illustrates the procedure.
2x 1 Example # 22 Resolve
2x 1
into partial fractions and evaluate
( x 1)( x 2) 2
2x 1 Solution.
Let
( x 1)( x 2 2)
=
2x 1 ( x 1)( x 2 2)
=
(x 1)(x
2
2)
dx
Bx C A + 2 . Then, x 1 x 2
A( x 2 2) (Bx C)( x 1) ( x 1)( x 2 2)
2x – 1 = A (x 2 + 2) + (Bx + C) (x + 1) ...(i) Putting x + 1 = 0 or, x = –1 in (i), we get – 3 = A(3) A = –1. Comparing coefficients of the like powers of x on both sides of (i), we get A + B = 0, C + 2A = –1 and C + B = 2 –1 + B = 0, C – 2 = –1 (Putting A = –1) B = 1, C = 1
2x 1
( x 1)( x 2) 2
=–
x 1 1 + 2 x 1 x 2
2x 1
Hence
(x 1)(x
2
2)
dx
= – n |x + 1| +
1 x 1 n |x 2 + 2| + tan–1 +C 2 2 2
CASE IV When some of the factors of the denominator g(x) are quadratic and repeating fractions of the
A (2ax b) A2 A A 0 (2ax b) 1 2 1 + 2 form 2 2 2 ax bx c ax bx c ax bx c ax bx c
2
A A 2k 2k 1( 2ax b ) + .......+ 2 k 2 ax bx c ax bx c
k
The following example illustrates the procedure.
2x 3 Example # 23 Resolve
( x 1)( x 2 1)2 2x 3
Solution.
Let
( x 1)( x 2 1)2
=
into partial fractions.
Dx E Bx C A + 2 + . Then, 2 ( x 1)2 x 1 x 1
2x – 3 = A(x 2 + 1)2 + (Bx + C) (x – 1) (x 2 + 1) + (Dx + E) (x – 1) Putting x = 1 in (i), we get – 1 = A (1 + 1) 2 A = –
......(i)
1 4
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MATHS Comparing coefficients of like powers of x on both side of (i), we have A + B = 0, C – B = 0, 2A + B – C + D = 0, C + E – B – D = 2 and A – C – E = –3.
1 and solving these equations, we get 4
Putting A = – B=
1 1 5 = C, D = and E = 4 2 2 2x 3
( x 1)( x 1) 2
2x
Example # 24 Resolve
Solution.
x 1 x5 1 + + 2 4( x 1) 4( x 1) 2( x 2 1)2
=
into partial fractions.
x3 1
2x
2x
We have,
2
=
x 1 3
( x 1)( x 2 x 1)
2x So, let
=
( x 1)( x x 1) 2
Bx C A + 2 . Then, x 1 x x 1
2x = A (x 2 + x + 1) + (Bx + C) (x – 1)
.......(i)
Putting x – 1 = 0 or, x = 1 in (i), we get 2 = 3 A A = Putting x = 0 in (i), we get A – C = 0 C = A =
2 3
2 3
Putting x = – 1 in (i), we get –2 = A + 2B – 2 C.
2 4 2 + 2B – B=– 3 3 3
–2= 2x
=
x 1 3
2 2x 2 2 1 1 (– 2 / 3) x 2 / 3 . + or 3 = + 2 3 3 x 1 3 x 1 x 1 x x 1
1 x x x 1 2
Self Practice Problems (7)
1
(i)
Evaluate :
(x 2)(x 3) dx
(ii)
Evaluate :
( x 1)(x
dx
Integration of type
ax
1)
x2 +C x3
n
Answers : (7) (i)
2
dx 2
bx c
,
(ii)
1 1 1 n |x + 1| – n (x 2 + 1) + tan–1 (x) + C 2 4 2
dx 2
ax bx c
,
ax 2 bx c dx
Express ax 2 + bx + c in the form of perfect square & then apply the standard results.
Example # 25 Evaluate : Solution.
x 2 2x 5 dx
We have,
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MATHS
x 2 2x 5
=
=
1 (x + 1) 2
( x 1)2 2 2 +
=
1 (x + 1) 2
x 2 2x 5 + 2 n |(x + 1) +
x 2 2x 1 4 dx =
Example # 26 Evaluate :
x
Solution.
1 2
x 1
x
1 2
x 1
x
dx =
2
( x 1)2 2 2 1 . (2)2 n |(x + 1) + 2
x 2 2x 5 | + C
dx 1 dx = 1 1 x 1 4 4
1
( x 1/ 2)
=
( x 1)2 2 2 | + C
2
3 /2
2
1
( x 1/ 2)
2
3/4
dx
x 1/ 2 +C tan–1 3 /2 3 /2 1
dx =
2x 1 + C. tan–1 3 3
2 =
Example # 27 Evaluate :
Solution.
1
dx
9 8x x 2
1 dx
9 8x x 2
=
1
{ x 8 x 9} 2
dx =
1
=
{( x 4)
2
5 } 2
dx =
1 { x 8 x 16 25} 2
1 5 ( x 4) 2
2
dx
x 4 +C dx = sin–1 5
Self Practice Problems (8)
Evaluate :
(9)
Evaluate :
Answers : (8)
2x
1 2
x 1
dx
1
2x 2 3x 2
1 n 3
2x 1 2x 2
dx
3 3 2 log x 4 x 2 x 1 + C 2
1 +C
(9)
Integration of type
ax
px q 2
bx c
Express
dx,
px q
2
ax bx c
dx,
(px q)
ax 2 bx c dx
px + q = A (differential coefficient of denominator) + B.
Example # 28 Evaluate :
2x 3 x 2 4x 1
dx
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MATHS Solution.
2x 3
dx
x 4x 1 2
=
=
=
( 2x 4) 1 dx
x 2 4x 1
2x 4 x 4x 1 2
dt t
–
dx – 1
( x 2) 2
t – n | (x + 2) +
=2
1 x 4x 1 2
3
Solution.
( x 5)
where t = (x 2 + 4x + 1) for Ist integral
dx,
2
x 2 4x 1 | + C
= 2 x 2 4 x 1 – n | x + 2 +
Example # 29 Evaluate :
dx
x 2 4x 1 | + C
x 2 x dx
d (x 2 + x) + . Then, dx x – 5 = (2x + 1) + .
Let (x – 5) = .
Comparing coefficients of like powers of x, we get 1 11 1 = 2 and + = – 5 = and = – 2 2 Hence,
( x 5)
1
x 2 x dx
11
=
2 (2x 1) 2
=
2 (2x 1)
1
1 = 2
x 2 x dx
x 2 x dx –
11 t dt – 2
11 2
2
x 2 x dx
1 1 x 2 2
2
dx
(where t = x 2 + x for first integral)
2 2 t3 / 2 1 11 1 x 1 x 1 1 = . – 2 2 2 2 3/2 2 2 2 2 2 x 1 x 1 1 1 1 – . n 2 2 2 + C 2 2 =
1 3/2 11 2x 1 x 2 x 1 n t – 8 3 2 4
1 2 x x x + C 2
=
1 2 11 2x 1 x 2 x 1 n (x + x)3/2 – 8 3 2 4
1 2 x x x + C 2
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14
MATHS Self Practice Problems
x 1
(10)
Evaluate :
x
(11)
Evaluate :
(12)
Evaluate :
( x 1)
2
x3
dx
6x 5 3x 2 5x 1
Answers : (10) Ans.
1 x x 2 dx 2x 1 + C tan–1 11 11 1
1 log |x 2 + x + 3| + 2 3x 2 5 x 1 + C
(11) Ans. 2 (12) Ans.
dx
1 2 3 (x + x + 1)3/2 – (2x + 1) 3 8
9 2 1 x x 2 – 16 log (2x +1 + 2 x x 1 ) + C
Integration of trigonometric functions (i)
(ii)
dx
OR
dx
2
a sin x b sinx cos x c cos 2 x
OR
dx a b sinx c cos x
a b cos x a b sin 2 x Multiply Nr & Dr by sec² x & put tan x = t.
dx a b sinx
OR
dx a b cosx
dx
OR
2
Convert sines & cosines into their respective tangents of half the angles and then, put tan
(iii)
a.cos x b.sinx c dx. .cos x m.sinx n
Express Nr A(Dr) + B
1
=
=
=
1
1 sin x cos x
dx
dx
1 1
2 tan x / 2 1 tan 2 x / 2 2 1 tan x / 2 1 tan2 x / 2
1 tan
Putting tan =
d (Dr) + C & proceed. dx
1 sin x cos x
Example # 30 Evaluate : Solution.
x =t 2
1
t 1
dx
1 tan 2 x / 2 2
x / 2 2 tan x / 2 1 tan x / 2 2
dx =
sec 2 x / 2 dx 2 2 tan x / 2
x 1 x = t and sec 2 dx = dt, we get 2 2 2
dt = n | t + 1| + C = n tan
x 1 + C 2
"manishkumarphysics.in"
15
MATHS Example # 31 Evaluate : Solution.
=
3 sin x 2 cos x
3 cos x 2 sin x
3 sin x 2 cos x
3 cos x 2 sin x
dx
dx
Let 3 sin x + 2 cos x = . (3 cos x + 2 sin x ) +
d (3 cos x + 2 sin x) dx
3 sin x + 2 cos x = (3 cos x + 2 sin x ) (–3 sin x + 2 cos x) Comparing the coefficients of sin x and cos x on both sides, we get =
12 5 and = – 13 13
=
(3 cos x 2 sin x ) (–3 sin x 2 cos x ) dx 3 cos x sin x
=
1 . dx +
=x+
dt
t
3 sin x 2 cos x
3 cos x 2 sin x
, where t = 3 cos x + 2 sin x 12 5 x– n | 3 cos x + 2 sin x | + C 13 13
= x + n | t | + C =
Example # 32 Evaluate : Solution.
dx
3 cos x 2
sin x 2 cos x 3
dx
We have, =
3 cos x 2
sin x 2 cos x 3
dx
Let 3 cos x + 2 = (sin x + 2 cos x + 3) + (cos x – 2 sin x) + Comparing the coefficients of sin x, cos x and constant term on both sides, we get – 2 = 0, 2 + = 3, 3 + = 2
=
6 3 8 ,= and = – 5 5 5
=
=
= x + log | sin x + 2 cos x + 3 | + 1
(sin x 2 cos x 3) (cos x 2 sin x ) dx sin x 2 cos x 3 cos x 2 sin x
dx sin x 2 cos x 3
where 1 =
1
sin x 2 cos x 3
Putting, sin x =
1 =
dx +
1
sin x 2 cos x 3
dx
dx
2 tan x / 2 1 tan 2 x / 2
, cos x =
1 tan 2 x / 2 1 tan 2 x / 2
1 2 tan x / 2 2(1 tan 2 x / 2) 3 1 tan2 x / 2 1 tan 2 x / 2
, we get
dx
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16
MATHS =
=
1 tan 2 x / 2
2 tan x / 2 2 2 tan tan
t
2
x / 2 2 tan x / 2 5
dx
dx
x 1 x = t and sec 2 = dt or 2 2 2
sec 2
2t 5
(t 1)
2
22
x tan 1 t 1 2 2 = tan–1 = tan–1 2 2 2
Hence, = x + log | sin x + 2 cos x + 3 | + tan–1
x tan 1 2 +C 2
6 3 8 ,= and = – 5 5 5
where =
dx
1 3 cos
Example # 33 Evaluate : Solution.
x dx = 2 dt, we get 2
2dt 2
dt
=2
x / 2 3(1 tan 2 x / 2)
sec 2 x / 2
Putting tan 1 =
2
2
x
Multiply Nr. & Dr. of given integral by sec 2x, we get =
sec 2 x dx tan x 4 2
=
tan x 1 +C tan–1 2 2
Self Practice Problems (13)
Evaluate :
Answer : (13)
Integration of type
4 sin x 5 cos x
5 sin x 4 cos x
dx
40 9 x+ log |5sinx + 4cosx| + C 41 41
sin
m
x.cosn x dx
Case - If m and n are even natural number then converts higher power into higher angles. Case - If at least one of m or n is odd natural number then if m is odd put cosx = t and vice-versa. Case - When m + n is a negative even integer then put tan x = t.
"manishkumarphysics.in"
17
MATHS Example # 34 Evaluate : Solution.
sin
Let =
sin
=–
5
Solution.
x cos 4 x dx
(1 t
(sin x)
put cos x = t
2 2
) . t4 . dt = –
1/ 3
Here m + n =
x cos 4 x dx
1 8
=
sin 1 8
1 = 16
=
2
(t
8
2t 6 t 4 ) dt
sin
2
(tan x )1/ 3
t
1/ 3
dt =
1 cos 2 x
dx
{put tanx = t sec 2x dx = dt}
3 4/3 3 t +C = (tanx)4/3 + C 4 4
x cos 4 x dx
2x(1 cos 2x )dx
sin
2t 2 1) t4 dt = –
1 7 – = – 2 (a negative integer) 3 3
=
Solution.
4
– sinx dx = dt
(cos x ) 7 / 3 dx
(sin x )1/ 3 (cos x ) 7 / 3 dx =
Example # 36 Evaluate :
(t
t9 t5 cos 9 x cos 5 x 2t 7 cos7 x + – +C =– +2 – +C 9 5 9 5 7 7
=–
Example # 35 Evaluate :
5
2
2x dx +
1 8
(1 cos 4 x ) dx
sin 1 16
2
2x cos 2x dx
sin3 2x 3
x sin 4 x sin 3 2x – + +C 16 64 48
Integration of type
x2 1 x 4 Kx 2 1
dx where K is any constant.
Divide Nr & Dr by x² & put x
1 = t. x
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18
MATHS Example # 37 Evaluate :
Solution.
Let =
Solution.
dx
1 x2 x4 1 x2
1 x x 2
= –
Example # 38 Evaluate :
1 x2
dt
x
dx
4
t 1
1 1
t 1 t 1
1 n 2
=–
2
4
=–
1 1 2 dx x 1 2 x 2 1 x
{put x +
1 1 = t 1 2 dx = dt} x x
1 1 x +C 1 x 1 x x
+C =–
1 n 2
dx
We have,
=
x
1 4
1
dx =
x
1 x2 2
1 = 2
1 = 2
Putting x –
=
1 x2
dx
x
2 x2 2
1 x2
dx
1 1 1 2 x2 x 1 dx = 1 1 2 x2 2 x2 2 x x
1 1 x2 dx – 2 2 1 x 2 x
1
1
1 1 x2 dx – 1 2 x2 2 x 1
1 x2 1 dx x2 2 x 1
1 x2 dx 2 1 x 2 x 1
1 1 = u in 1st integral and x + = in 2nd integral, we get x x du
u 2
1 2
1 = 2
2
2
–
1 2
d
2 2
2
u 2 1 – 1 tan–1 n +C 2 2 2 2 2 2 2 1
=
x 1/ x x 1/ x 2 1 – tan–1 n +C x 1/ x 2 2 2 4 2 2 1
=
1 =
2 2
tan
–1
x 2 1 x2 2 x 1 1 2 x – 4 2 n x 2 x 2 1 + C
"manishkumarphysics.in"
19
MATHS Self Practice Problem :
x2 1
(14)
Evaluate :
x 4 7x 2 1
(15)
Evaluate :
tan x dx
dx
1 3 x 1 x 3 x x
Answers :
1 n 6
(14)
+C
y 1 + tan–1 n 2 2 2 2
1 (15)
y 2 + C where y =
y 2
tan x –
1 tan x
Integration of type
(ax
dx b) px q
dx
ax
OR
2
bx c px q
.
Put px + q = t2.
1
( x 3)
Example # 39 Evaluate : Solution.
( x 3)
=
=2
Let =
dx
1
Let =
(t
(x
x 1
{Put x + 1 = t2 dx = 2t dt}
dx
1 2
t
(x
Example # 40 Evaluate :
Solution.
x 1
2t
dt
1 3) t 2 dt 2
2
2
t2 1 n + C t2 2( 2)
=2.
2
1 n 2
x 1 2 x 1 2
+ C.
x2 2
3 x 3) x 1
dx
x2 2
3 x 3) x 1
dx
Putting x + 1 = t2, and dx = 2t dt, we get =
=
t
( t 2 1) 4
t2 1
dt = 2
1 t2 dt 1 t2 2 1 t
( t 2 1) 2t dt 2
1)2 3( t 2 1) 3} t 2
1
du
=2
{(t
u 3 2
2
{put t –
u 2 +C= tan–1 tan–1 3 3 3
2 =
1 = u} t
"manishkumarphysics.in"
1 t t 3 +C
20
MATHS 2
=
3
tan
–1
t 2 1 2 –1 t 3 + C = 3 tan
x +C 3 ( x 1)
Integration of type
(ax (ax
dx b) px
2
qx r
2
,
2
b) px q
put x =
1 ; t
1 t
dx
(x 1 ) x
2
x 1
dx
Let =
(x 1 ) x
=
Example # 42 Evaluate
2
dt 2
1 t2 t
1 1 1 t t
1 t
=
=–
dt t
= 1 1 1 t2 t
dt t t 1 2
1 1 3 t 2 2 4
+ C, where t =
1 x 1
dx 2
) 1 x 2
dx = –
(t
1 1 dx = – 2 dt } t t
2
1 3 t 2 4
(1 x
=
= – n t
2
Put x =
{put x + 1 =
x 1
dt
=
Solution.
put ax + b =
dx
Example # 41 Evaluate
Solution
,
1
dt
t2
dt 2
{put
1) t 2 1
(y
y dy 2
2) y
t2 – 1 = y2 tdt = ydy }
y + C = – 1 tan–1 tan–1 2 2 2
1 =–
1 x2 2x
+C
Self Practice Problems : dx
(16)
Evaluate :
( x 2) x 1
(17)
Evaluate :
( x 5 x 6) x 1
(18)
Evaluate :
dx 2
dx ( x 1) 1 x x 2
"manishkumarphysics.in"
21
MATHS dx
(19)
Evaluate :
(2 x 2 1) 1 x 2
(20)
Evaluate :
( x 2 x 2) x 2 2 x 4
dx 2
2 tan–1
Answers : (16)
x 1 + C
(17) 2 tan–1
3 1 2 x 1 +C 5 2
1 x2 tan–1 3 3 x
1
(18)
sin–1
(20)
x 2 2x 4 6 ( x 1) – n 2 +C 2 6 x 2x 4 6 ( x 1)
(19) –
x 1 –1 +C 2 tan 2
x 1 –
+C
1
Integration of type
xα β x dx or
xα β x
dx xα xβ
dx or
x α β
2.
3.
n =
tan
n
put x =
cos2 + sin2
x α x β dx;
put x =
sec2 tan2
;
put x
tan
n
Reduction formula of 1.
dx;
tan
x dx =
2
x
= t2 or x = t2.
x dx ,
x tann2 x dx =
n =
sec
n =
tann1 x – n – 2 , n 2 n 1
n =
cot
n
x dx =
n =
n = –
sec
n
x tann2 x dx
cot
cos ec
n =
2
2
2
cot
n
x dx ,
(sec
2
sec
n
x dx ,
cos ec x dx n
x 1) tann – 2x dx
– n – 2
x . cot n2 x dx =
(cos ec x 1) cot 2
n 2
x dx
x cot n2 x dx – n – 2
cot n1 x – n – 2 , n 2 n 1
x dx =
sec
2
x sec n 2 x dx
n = tanx sec n – 2x –
(tan x)(n 2) sec
n–3
x. secx tanx dx.
"manishkumarphysics.in"
22
MATHS n = tanx sec n – 2 x – (n – 2)
(n – 1) n = tanx sec n – 2x + (n – 2) n – 2 n =
4.
(sec
n =
2
x – 1) sec n – 2x dx
n2 tan x sec n2 x + n 1 n– 2 n 1
cos ec x dx = cos ec x cosec n
2
n–2
x dx
(cot x)(n 2) (– cosec
n = – cotx cosec n – 2x +
– cotx cosec n – 2x – (n – 2)
n = – cotx cosec n – 2x – (n – 2)
(n – 1) n = – cotx cosec n – 2 x + (n – 2) n – 2 n =
cot
2
n =
x cosec x cot x) dx
x cos ec n2 x dx
(cos ec
2
x 1) cosec n – 2 x dx
cot x cos ec n 2 x n2 + – n 1 n 1 n– 2
Example # 43 Obtain reducation formula for n = Solution.
n–3
(sin x) (sin x)
n –1
sin x dx. Hence evaluate sin x dx n
4
dx
(sin x)
= – cos x (sin x)n–1 + (n – 1) = – cos x (sin x)n–1 + (n – 1)
(sin x)
n–2
n–2
cos 2x dx (1 – sin2x) dx
n = – cos x (sin x)n–1 + (n – 1) n–2 – (n – 1) n
n = –
(n 1) cos x(sin x )n1 + n–2 n n
(n 2)
3 cos x(sin x ) 1 x cos x(sin x )3 +C + 2 2 4 4
Hence 4 = –
Self Practice Problems :
x 3 dx x4
(21)
Evaluate :
(22)
Evaluate :
[( x 1)(2 x )]3 / 2
(23)
Evaluate :
[( x 2) ( x 1)6 ]1/ 7
(24)
Deduce the reduction formula for n =
(25)
If
dx
dx 8
m,n =
(sin x)
m
dx (1 x )
4 n
and Hence evaluate 2 =
dx (1 x 4 )2
.
(cos x)n dx then prove that
"manishkumarphysics.in"
23
MATHS m,n =
Answers :
n 1 (sin x )m1(cos x )n1 + . m n m,n–2 mn
(21)
( x 3)( x 4) + n
(22)
x 1 2 x 2 2 x x 1 + C
(23)
7 x 1 3 x 2
(24)
n =
x3 x4 + C
1/ 7
+C
x 4(n 1)(1 x 4 )n 1
+
4n 5 4(n 1) n–1
1 x x 3 1 1 1 x tan n 2 = + 4 (1 x 4 ) 4 2 2 2 4 2 x x
"manishkumarphysics.in"
1 2 x + C 1 2 x
24
