INTEGRALS
93
5.3 (i)
1 cot ax + c. If a = 1, then a
13.
cosec
14.
15.
a
16.
17.
a
18.
x
19.
21.
x
23.
x a 2 x 2 dx = 2
x a2 –1 a 2 x 2 + 2 sin a + c.
24.
x a 2 x 2 dx = 2
a2 2 2 a 2 x 2 + 2 log x a x + c.
25.
x x 2 a 2 dx = 2
a2 2 2 x 2 a 2 – 2 log x x a + c.
2
ax dx = –
eax + c. If a = 1, then a
eax dx =
1
x
ax a dx = + c. loge a x
x + c. If a = 1, then a
1 1 x2
1 1 x dx = tan–1 + c. If a = 1, then x2 a a
1 x
a x
2
dx = sin–1
1 x a 2
1 2
a x
2
2
dx =
x dx = –cot x + c.
2
+ c.
2
2
x
amx dx = + c. If m = 1, then m loge a
mx
2
e dx = e
cosec
1
1 x sec–1 + c. If a = 1, then a a
2 2 dx = log x a x + c.
1 xa 1 log + c. 2 dx = 2 a x a a
dx = sin–1x + c.
2
dx = tan–1x + c. 1
x
x2 1
dx = sec–1x + c. 1
20.
22.
a
x a2 2
2
2 2 dx= log x x a + c.
1 ax 1 log + c. 2 dx = 2 a ax x
METHODS OF INTEGRATION Integration by Substitution (or change of independent variable) :
f(x) dx be changed to t, then we substitute x = (t) f(x) dx = f((t))’(t) dt
If the independent variable x in i.e., dx = ‘(t) dt (ii)
which is either a standard form or is easier to integrate. Integration by parts : If u and v are the differentiable functions of x then
u. v dx = u vdx – dx u vdx dx. d
How to choose Ist and IInd functions: If the two functions are of different types take that function as Ist which comes first in the word ILATE where I stands for inverse circular function, L stands for logarithrmic function A stands for Algebratic function, T stands for trigonometric function and E stands for exponential function. (b) If both functions are algebratic take that function as Ist whose differential coefficient is simple. (c) It both functions are trigonometrical take that function as IInd whose integral is simpler. (d) Successive integration by parts : Use the following formula (a)
uvdx = uv n
1
– u’v2 + u”v3 – u’” v4 + ......... .........+ (–1)
n–1
u
n –1
vn + (–1)
n
u
n
vndx.
where u stands for nth differential coefficient of u w. r. t. x and vn stands for n th integral of v w. r. t. x. cancellation of Integrals : i.e.
e {(f(x) + f’(x)} dx = e f(x) + c. x
x
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INTEGRALS
95
find l and m by comparing the coefficients of sin x and cos x and split the integral into sum of two integrals as l
d.c. of Dr r = lx + m ln | D | + c Dr
dx + m
Rule for (v) : Express the numberator as r r l(D ) + m (d.c. of D ) + n, find l, m, and n by comparing the coefficients of sin x, cos x and constant term and split the integral into sum of three integrals as l
dx d.c. of Dr r dx + n = lx + m ln | D | + n r Dr D
dx + m
dx
D
r
and to evaluate
dx
D
r
proceed by
the method to evaluate rule (i), (ii) and (iii).
Type V : Integrals of the form
x
a 2 dx x kx 2 a 4
(i)
2
4
x
a 2 dx x kx 2 a 4
(ii)
2
4
where k is a constant, + ve, –ve or zero. Rule for (i) and (ii) : Divide above and below by x
2
then putting (i) t = x –
a2 x
a2 a2 dt = 1 x 2 dx and dt = 1 x 2 dx
i.e.,
then the questions shall reduce to the form
t
2
dt or c2
t
2
dt c2
Remember : (i)
x 2dx 1 4 2 4 = x kx a 2 dx
(ii)
x 4 kx2 a4
(iii)
x
dx 2
k
=
n
=
x
a 2 dx 1 + x kx 2 x 4 2
2
x
x
x k 2n 2 x k 2
2
4
n1
+
a 2 dx x kx 2 a 4 2
4
a 2 dx 1 – x kx 2 x 4 2a 2
1 2a 2
4
x
a 2 dx x kx 2 a 4
2
4
dx 2n 3 n1 2 k 2n 2 x k
Type VI : Integrals of the form (i)
dx
Ax B
(iv)
Ax
(ii)
ax b dx
2
Bx c
ax
2
bx c
Rule for (i) and (ii) : Rule for (iii) :
Bx C
Put Ax + B = Put
px q
x ax b , a b
(iii)
ax b
Ax B ax dx
2
bx c
2
1 t
ax 2 bx c Ax 2 Bx C
Integration by partial fraction : Form of the Rational Function 1.
Put ax + b = t
Rule for (iv) : (iv)
Ax
dx 2
=t
2
Form of the Partial Fraction
A B x a x b
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INTEGRALS
97
SOLVED PROBLESM Ex.1
Sol.
Find f(x), if 3 (i) f’(x) = (x – 1) and f(2) + f(–2) = 0 (ii) f’(x) = a (cos x + sin x ), f(0)
y dy =
= 9 and f = 15 2
1 (ii) Put x + log x = y. Then, 1 dx = dy x
(i)
Here, f ’(x)
f(x)
= (x – 1)
3
= (x–1) dx =
x 1 dx = dy x
1 4 (x – 1) + 4
C f(2)
=
I=
1 1 4 (2 – 1) + C = = + C 4 4
1 81 +C+ +C=0 4 4
41] (ii)
x log x 3
82 + 2C 4
(iii)
(y
3 2
I=
f(x) =a
=
and
f =15 = a×1 – a × 0 + C 2
Integrate the following functions w.r.t. x: (i)
1 log x 2 (ii) x 1 x log x x
2
x
(i) Put 1 + log x = y. Then,
So, I =
1 log x 2 x
1 dx = dy x
1 2
dy
1
dy 2)y 2 dy 2
=
Find : (i)
Sol.
(i)
cos x dx (ii) 3
We have
+ 3 cos x)
1 sin x dx
3
cos x =
cos
3
1 (cos 3x 4
x dx =
1 4
3
cos 3x dx + 4 cos xdx =
1 sin 3 x 3 1 3 + sinx + C = sin3x + sinx 4 3 4 12 4 +C (ii)
We have
1 sin x
(iii) x x 2 Sol.
3 2
y 2y
2 2 2 y 2. y 3 C 5 3
Ex.3
a+C =15 From (2) and (3), we have a = 3, C = 12 Hence, from (1), we have f(x)=3(sin x–cos x)+12 Ex.2
y3 + C = 3
=
5 3 2 x 22 4 x 22 C 5 3
sin x dx
= a sin x – a cos x + C ...(1) f(0) = 9 = a×0 – a×1+C ...(2) C– a=9
(y
5
a (cos x + sin x)dx +a
dy
x 2 dy =
1
= a (cos x + sin x)
cos x dx
x
2y 2 ) dy =
2
Put x + 2 = y. Then, dx = dy So,
y
+C
3
1 1 4 4 (x – 1) – 10.25 = [(x – 1) – 4 4 Here, f ’(x)
x
=
=0 20.5 + 2C = 0 C = – 10.25 Hence, from (1) we have f(x) =
x 1x log x 2 dx
Given that f(2) + f(–2) = 0, we have
+C
3
So,
y3 1 log x +C= 3 3
3
2
x x x 2 x sin2 2 sin cos cos 2 2 2 2
dx
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=
INTEGRALS
99 1 1 1 = + ( x 1)( x 2) x 1 x2
Thus,
dx
( x 1)(x 2)
1
=
Sol.
=
Sol.
x 1 +C x2
Find (ii)
:
(x - 1)2 )(x + 2) Let
( x 2 )2 dx
3x 1
1 2
3
7
=
2 ; 9
B=
1 ; 3
( x 1)2 ( x 2)
2
2
1 1 sin 2t t . cos 2t . C 4 4 2
1 1 t . (1 2 sin2 t ) sin t 1 sin2 t C 4 4 1 1 (2x 2 1) sin1 x x 1 x 2 C 4 4 Put x = cos t so that dx = – sin t So,
(ii) dt
x cos
1
x dx cos t . t . sin t dt =
C
=
2 1 2 9 dx 3 9 dx dx x 1 x2 ( x 1)2
1 t sin 2t dt 2
1 1 t(2 cos2 t – 1) – 4 4
t+C= Ex.9
Sol.
dx
1 1 2t t . cos 2t . sin C 4 4 2 give below)
Find : (i)
(i) = dt 1
=
(Using (1)
1 cos2 t . cos
1 x (2x2 – 1) cos–1 x – 4 4
=
2 2 1 log |x – 1| – – log |x + 2| 9 9 3( x 1) +C
=
(ii)
x dx sin t . t . cos t dt
t . sin 2t dt
2 – 9
=
1
=
= 3
C x2 Then x = A (x – 1) (x + 2) + B(x + 2) + C (x + 1)2 x = A (x2 + x – 2) + B (x + 2) + C (x2 – 2x + 1) Comparing coefficients of x2, x and the constant terms on both sides, we get 0 = A + C; 1 = A + B – 2C ; 0 = –2A + 2B +C
x
x sin
1 t . sin 2t dt 1. sin 2t dt dt 2
=–
+
Thus,
x dx
So,
2
B A + ( x 2) 2 x2
B A x Let + 2 ( x 1)2 x 1 ( x 1) ( x 2)
A=
-1
1 cos 2t cos 2t = t . dt
1 log |x + 2| – 7 . +C ( x 2)
x cos
x dx
dt
dx
( x 2)2 dx x 2 dx ( x 2)2 dx
(ii)
-1
Put x = sin t so that d x = cos t .
=
( x 2) 2 Then, 3x – 1 A(x + 2) + B Comparing coefficients of x and the constant terms on both sides, we get 3 = A and –1 = 2A + B A = 3 and B = –7 Thus, 3x 1
x sin
(i)
=
3x 1
(i)
x
(i)
Find : (i) (ii)
= log | x+1 | – log | x+2 | + C
Ex.7
Ex.8
1
x 1 dx x 2 dx log
x 1 1 2 log – +C x2 3( x 1) 9
=
1 x 2 + C
1 1 dt 2 logx (logx)
e
x
sin4x - 4 1 - cos4x dx
Let log x = t. Then x = et and
1
1
log x (log x)2 dx
dx x =
1 t dx
t t 2 e
t = e [ f ( t ) f ' ( t ) dt, where f ( t )
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1 = et . f(t)+C t
INTEGRALS
101
UNSOLVED PROBLEMS EXERCISE – I cos ec x dx cos ec x cot x
Q.1.
Evalute :
(i)
Q.2
Evalute :
(i)
tan
Q.3
(i) If f(x) = sin2x + cos3x + 5, f(0) =
1
sin 2x dx 1 cos 2x
sin6 x cos 6 x dx sin2 x. cos 2 x
(ii)
(ii)
tan
1
(iii)
(sec x tan x ) dx , –
1 cos 3 x dx
x 2 2
3 , find (x) 2
(ii) If f(x) = a sinx + b cos x and f(0) = 4, f(0) = 3, f = 5, find f(x) 2
(iii) The gradient of a curve is given by 2x
3 x2
. The curve passes through (1,1). Find its
equation. (iv) The gradient of a curve is 6x2 – 2ax + a2. The curve passes through the point (0, 0) and (1, 8). Find a. Q.4
Evaluate :
(i)
Q.5
Evaluate :
(i)
sin x sin 2x sin 3x dx
sin(x a) dx sin x
(ii)
tan x tan 2x tan 3x dx
(ii)
sin( x a) dx
(ii)
x
(ii)
sin x
(
i
i
1
sin( x a) sin( x b)dx e
1 dx ex
Q.6
Evalute :
Q.7
Evaluate :
Q.8
Evalute :
Q.9
Evaluate :
(i)
Q.10
Evaluate :
(i)
sin (log x) dx
Q.11
Evaluate :
Q.12
Evaluate :
(i)
Q.13
Evaluate :
(i)
( 3 x 2)
Q.14
Evaluate :
(i)
3 4 cos x dx
Q.15
Evaluate :
(i)
1 cot x dx
Q.16
Evaluate :
(i)
(x
(i)
(i)
sin 2x
(a b cos x) dx 2
(i)
x
1 x
1 dx x 1/ 3
1 sin x sin( x a) 3
log (1 x )(1 x ) dx
(ii)
(ii)
sec 3 x dx
1 x
1/ 2
dx
sin 1 x dx x2
sin 1 x cos 1 x
sin
1
x cos 1 x
(ii)
, a n, n z
e
ax
dx
cos bx dx
dx
sin ( x ) dx sin ( x )
x 2 x 1 dx
1
1
2
1 dx 1) ( x 2 4)
(ii)
ax dx ax
(ii)
( 2x 5)
(ii)
1 cos x sin x dx (iii) 3 2 sin x cos x dx
(ii)
3 cos x 2 sin x dx
(ii)
(x
x 2 4 x 3 dx
cos x
1
3 sin x 2 cos x
2
2x dx (iii) 1) ( x 2 3)
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x2 2 dx ( x 1) ( x 2 4) 2
i
)
INTEGRALS
103
BOARD PROBLES EXERCISE – II /2
Q.1
(i)
Evaluate
3
sin7 x
dx .
sin7 x cos7 x
0
(x
(ii)
2
2x ) dx as limit of a sum.
[C.B.S.E.
0
2000]
Q.2
2x 1
(i)
2
x 4x 3
sin
4
dx (ii)
x dx .
(iii)
1 sin 2x
x cos2 x dx
x sin
(iv)
1
x dx .
[C.B.S.E. 2000]
Q.3
sin 7x sin x dx .
(i)
(ii)
4x 3 2
2x 2x 3
log (1 x
(iii)
dx
2
) dx .
(iv)
x 1
(x 1)(x 2) dx .[C.B.S.E. 2001]
2
Q.4
(i) Evaluate as the limit of a sum
2
( x 3) dx . (ii) Evaluate
x x sin x dx . (iii) dx . 1 sin x 1 cos2 x 0
0
0
[C.B.S.E. 2001]
Q.5
e
(i)
ax
cos bx dx . (ii)
x
( x 2 1)( x 1) dx
(iii)
dx x 1 x 2
(iv)
dx
x2 4x 8 .[C.B.S.E.
2002] /2
/2
Q.6
(i)
Evaluate
cos 2 x dx .
(ii)
Prove that
/4
tan x cot x ) dx 2 .
0
0
(iii) Prove
(
log (1 tan ) d
0
log 2. (iv) 8
2
Evaluate
e 1
x
1 1 x 2 dx . x
[C.B.S.E. 2002] Q.7
(i)
(iv)
sec
3
(ii)
x
tan 1 x
(1 x)2 dx .
(iii)
x dx
(x 2)(3 2x) .
(i)
a
sin 2x . log (tan x ) dx 0.
(ii)
Prove that
/2
/2
Evaluate
a
0
(iii)
cos 2x . log sin x dx .
/4
1
1 sin x dx. 1 sin x
[C.B.S.E. 2003]
/2
Q.8
tan
(iv)
0
ax dx = a. ax
sin2 x dx sin x cos x
[C.B.S.E. 2003]
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INTEGRALS
105 Q.19
(i)
x sin
1
sec 2 x dx 3 tan x
(ii)
ex 5 4e x e2x
(iii)
dx
( x 4) e x ( x 2)3
dx (iv)
x dx
[C.B.S.E. 2009]
Q.20
(i)
ecos x
ecos x ecos x
/2
dx (ii)
(2 log sin x log sin 2x ) dx (iii)
0
0
0
[C.B.S.E. 2009] Q.21
(i)
sec
2
(ii)
(7 4 x ) dx
(i)
(x 2) (x 3)
[C.B.S.E. 2010]
/2
Q.22
x2
2
sin5 x dx
(ii)
/ 2
dx
5x 2
x2 4x 3 dx 1
[C.B.S.E. 2010] Q.23
6x 7 (x 5)(x 4)
dx
[C.B.S.E. 2011] /2
Q.24
Evaluate :
(i)
[C.B.S.E. 2011] 2
Q.25
Evaluate (i)
1
0
/3
x sin x dx 1 cos x
/6
x3 x dx .
(ii)
x sin x
(ii)
1 cos
(ii)
x 1 dx 2 (x 1) (x 3)
2
0
x
dx 1 tan x
dx
[C.B.S.E. 2012] Q.26
Evaluate (i)
1
x sin x 1x
2
dx
2
[C.B.S.E. 2012] Q.27
Evaluate : [C.B.S.E. 2013] cos 2x cos 2 dx cos x cos Evaluate : x2 dx 2 x 2x 3
Q.28
Evaluate : [C.B.S.E. 2013]
Q.29
dx
xx
5
3
Evaluate : 2
[C.B.S.E. 2013]
1
1e 0
sin x
x dx
a2 cos2 x b2 sin2 x dx
dx
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INTEGRALS
107 x2 1 2 1 log x 2 x 1 c tan 1 2 2 2 x 2x 1 2x 4 2 1
x 2 1 tan 1 c 3 3 x
1
(ii)
(iii)
tan x 1 tan1 c 2 2 tan x
1
18. (i)
5 1 1 e2 e (ii) 1 (iii) (iv) log 2ab(a b) 2 2 2 2( 5 2) 2
19.
(i) cos a–cos b (ii)
1 15 e8 (iii) 14 (iv) 36 21. (i) – (ii) 0 4 2 2
23. (i) 4 (ii) 2e–2 (iii) 4 (iv)
19 2
24. (i) 66 (ii)
2 1 (iii) 1 (iv) log 2 1 4 2 2
(ii)
1 (n 1)(n 2)
(v)
2 4
13 10
(iii) 2
26. (i)
4
22.
12
(iv) 4 25.
(ii)
4
(i) –
1 2
log ( 2 – 1)
log 2 (iv) 8
(iii)
0
( v)
(vi) 0
EXERCISE – 2 (BOARD PROBLEMS)
1. (i)
(ii) 18 4
2 2 2. (i) 2 x 4 x 3 3 log( x 2) x 4 x 3 c
(ii)
1 [12x – 8 sin 2x+sin 32
4x]+c 2
(iii) log (x + cos x)+c (iv)
x x2 –1 sin x + 4 2
1 –1 1 x 2 – 4 sin x + c
2 16. (i) 1 (ii) (iii) log ( 2 + 1) (iv) – log 2 2 2 2 4 2 (i)
3 20 (ii) (iii) (iv) 5 2 2 4 3
(ii) 2 2x 2 2x 3
2
3.
(i)
(iii)
eax 2
a b
2
17. (i)
93 2
(ii) /12
(ii)
16 2 (iii) 15 4
2
18.
sin 6 x sin 8 x +c 12 16
1 3 log x x 2 x 2 2 2
1
–1
(iii) x log (1 + x ) – 2(x – tan x) + c (iv)
5. (i)
13. (i) /2
2 1 log(x + 1) + log (x – 2) + c 3 3
[a cos bx + b sin bx] + c
2 2 3/2 3/2 log (x + 2) – log(x + 1) +c 3 3
7. (i) sec x tan x + log (secx + tan x) (ii)
(ii)
(iv)
4. (i)
26 2 (ii) (iii) 3 4
1 1 1 2 –1 log (x + 1) + tan x – log (x + 1) 4 2 2
1 -1 x 2 tan 6. (i) 2 2 4
(iv)
e2 e 2
1 x x 2 tan 1 x 1 –1 2 + [log (1 + x) + tan x – log (1 + x )] + c (iii) – +c 2 2 4 1 x 4
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