1. If f & g are functions of x such that g′ (x) = f(x) then, ∫ f(x) dx = g(x) + c ⇔ d dx {g(x)+c} = f(x), where c is called the constant of integration. 2. 2. 2. 2. Standard Formula: Standard Formula: Standard Formula: Standard Formula: (i) ∫ (ax + b) n dx = ( ) ( ) ax b an n + + +1 1 + c, n ≠ -1 (ii) ∫ dx ax b + = 1 a ln (ax + b) + c (iii) ∫ e ax+b dx = 1 a e ax+b + c (iv) ∫ a px+q dx = 1 p a na px q + + c; a > 0 (v) ∫ sin (ax + b) dx = - 1 a cos (ax + b) + c (vi) ∫ cos (ax + b) dx = 1 a sin (ax + b) + c (vii) ∫ tan(ax + b) dx = 1 a ln sec (ax + b) + c (viii) ∫ cot(ax + b) dx = 1 a ln sin(ax + b)+ c (ix) ∫ sec² (ax + b) dx = 1 a tan(ax + b) + c (x) ∫ cosec²(ax + b) dx = - 1 a cot(ax + b)+ c (xi) ∫ sec (ax + b). tan (ax + b) dx = 1 a sec (ax + b) + c (xii) ∫ cosec (ax + b). cot (ax + b) dx = - 1 a cosec (ax + b) + c (xiii) ∫ secx dx = ln (secx + tanx) + c OR ln tan π 4 2 + x + c (xiv) ∫ cosec x dx = ln (cosecx - cotx) + c OR ln tan x 2
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1. If f & g are functions of x such that g′(x) = f(x) then,
∫ f(x) dx = g(x) + c ⇔ d
dx{g(x)+c} = f(x), where c is called the constant of integration.
2 .2 .2 .2 . Standard Formula:Standard Formula:Standard Formula:Standard Formula:
(i) ∫ (ax + b)n dx =( )
( )ax b
a n
n+
+
+1
1 + c, n ≠ −1
(ii) ∫ dx
ax b+ =
1
a ln (ax + b) + c
(iii) ∫ eax+b dx =1
a eax+b + c
(iv) ∫ apx+q dx =1
p
a
na
px q+
� + c; a > 0
(v) ∫ sin (ax + b) dx = −1
a cos (ax + b) + c
(vi) ∫ cos (ax + b) dx =1
a sin (ax + b) + c
(vii) ∫ tan(ax + b) dx =1
a ln sec (ax + b) + c
(viii) ∫ cot(ax + b) dx =1
a ln sin(ax + b)+ c
(ix) ∫ sec² (ax + b) dx =1
a tan(ax + b) + c
(x) ∫ cosec²(ax + b) dx = −1
acot(ax + b)+ c
(xi) ∫ sec (ax + b). tan (ax + b) dx =1
a sec (ax + b) + c
(xii) ∫ cosec (ax + b). cot (ax + b) dx = −1
a cosec (ax + b) + c
(xiii) ∫ secx dx = ln (secx + tanx) + c OR ln tan π
4 2+
x+ c
(xiv) ∫ cosec x dx = ln (cosecx − cotx) + c OR ln tanx
2
(xv) ∫d x
a x2 2− = sin−1
x
a + c
(xvi) ∫d x
a x2 2+
=1
a tan−1
x
a + c
(xvii) ∫d x
x x a2 2− =
1
a sec−1
x
a + c
(xviii) ∫d x
x a2 2+
= ln [ ]x x a+ +2 2 OR sinh−1
x
a + c
(xix) ∫d x
x a2 2− = ln [ ]x x a+ −2 2
OR cosh−1x
a + c
(xx) ∫d x
a x2 2−
=1
2a ln
xa
xa
−
+ + c
(xxi) ∫d x
x a2 2− =
1
2a ln
ax
ax
+
− + c
(xxii) ∫ a x2 2− dx =x
2a x2 2− +
a2
2 sin−1
x
a + c
(xxiii) ∫ x a2 2+ dx =x
2x a2 2+ +
a2
2 �n
++
a
axx 22
+ c
(xxiv) ∫ x a2 2− dx =x
2x a2 2− −
a2
2 �n
−+
a
axx 22
+ c
(xxv) ∫ eax. sin bx dx =e
a b
ax
2 2+ (a sin bx − b cos bx) + c
(xxvi) ∫ eax. cos bx dx =e
a b
ax
2 2+ (a cos bx + b sin bx) + c
3 .3 .3 .3 . Theorems on integrat ionTheorems on integrat ionTheorems on integrat ionTheorems on integrat ion
(i) ∫ dx).x(fc = c ∫ dx).x(f
(ii) ∫ ± dx))x(g)x(f( = ∫ ± dx)x(gdx)x(f
(iii) ∫ += c)x(gdx)x(f ⇒ ∫ + dx)bax(f = a
)bax(g + + c
Note : (i) every contineous function is integrable
(ii) the integral of a function reffered only by a constant.
∫ dx).x(f = g(x) + c
= h(x) + c
g′(x) = f(x) & h′(x) = f(x)
g′(x) – h′(x) = 0
means, g(x) – h(x) = c
Example : Evaluate : ∫ dxx4 5
Solution. ∫ dxx4 5 =
6
4 x6 + C =
3
2 x6 + C.
Example : Evaluate : ∫
++−+ dx
x
2
x
74x5x 23
Solution. ∫
++−+ dx
x
2
x
74x5x 23
= ∫ dxx3 + ∫ dxx5 2
– ∫ dx4 + ∫ dxx
7 + ∫ dx
x
2
= ∫ dxx3 + 5 . ∫ dxx2
– 4 . ∫ dx.1 + 7 . ∫ dxx
1 + 2 . ∫
− dxx 2/1
= 4
x4
+ 5 . 3
x3
– 4x + 7 log | x | + 2
2/1
x 2/1
+ C
= 4
x4
+ 3
5x3 – 4x + 7 log | x | + 4 x + C
Example : Evaluate : ∫ ++ alogaxlogaalogx eee dx
Solution. We have,
∫ ++ alogaxlogaalogx eee dx
= ∫ ++aax alogxlogalog eee dx
= ∫ ++ )axa( aax dx
= ∫ dxax + ∫ dxxa
+ ∫ dxaa
= alog
ax
+ 1a
x 1a
+
+
+ aa . x + C.
Example : Evaluate : ∫+x
xx
5
32 dx
Solution. ∫+x
xx
5
32 dx
= ∫
+
x
x
x
x
5
3
5
2 dx
= ∫
+
xx
5
3
5
2 dx =
5/2log
)5/2(
e
x
+ 5/3log
)5/3(
e
x
+ C
Example: Evaluate : ∫ xcosxsin 33 dx
Solution. = 8
1
3)xcosxsin2(∫ dx
= 8
1 ∫ x2sin3 dx
= 8
1 ∫
−
4
x6sinx2sin3 dx
= 32
1 ∫ − )x6sinx2sin3( dx
= 32
1
+− x6cos
6
1x2cos
2
3 + C
Example : Evaluate : ∫ +1x
x2
4
dx
Solution. ∫ +1x
x2
4
dx
= ∫ +
+−
1x
11x2
4
dx = ∫ +
−
1x
1x2
4
+ 1x
12 +
dx
= )1x( 2 −∫ dx + ∫ +1x
12 dx =
3
x3
– x + tan–1 x + C
Example: Evaluate : ∫ + 2x94
1 dx
Solution. We have
∫ + 2x94
1
= 9
1 ∫
+ 2x9
4
1 dx
= 9
1 ∫ + 22 x)3/2(
1 dx
= 9
1 .
)3/2(
1 tan–1
3/2
x + C =
6
1 tan–1
2
x3 + C
Example : ∫ dxx2cosxcos
Solution. ∫ dxx2cosxcos
= 2
1∫ dxx2cosxcos2
= 2
1∫ + )xcosx3(cos dx
= 2
1
+
1
xsin
3
x3sin + c
Self Practice Problems
1. Evaluate : ∫ xtan2 dx Ans. tanx – x + C
2. Evaluate : ∫ + xsin1
1 dx Ans. tanx – sec x + C
4.4 .4 .4 . Integration by SubsitutionsIntegration by SubsitutionsIntegration by SubsitutionsIntegration by Subsitutions
If we subsitute x = φ(t) in a integral then(i) everywhere x will be replaced in terms of t.(ii) dx also gets converted in terms of dt.(iii) φ(t) should be able to take all possible value that x can take.
Example : Evaluate : ∫43 xsinx dx
Solution. We have
Ι = ∫43 xsinx dx
L e t xL e t xL e t xL e t x
4 = t ⇒ d(x4) = dt ⇒ 4x3 dx = dt ⇒ dx = 3x4
1 dt
Example : ∫ x
)xn( 2�
dx
Solution. ∫ x
)xn( 2�
dx
Put �nx = t
⇒x
1 dx = dt
= ∫
x
dx.t2
= ∫ dtt2
= 3
t3
+ c
)xn( 3
Example : Evaluate ∫ + dxxcos)xsin1( 2
Solution. Put sinx = tcosx dx = dt
∫ + dt)t1( 2 = t +
3
t3
+ c
� + c=
3
= sin x + 3
xsin3
+ c
Example : Evaluate : ∫ ++ 1xx
x24 dx
Solution. We have,
Ι = ∫ ++ 1xx
x24 dx = ∫ ++ 1x)x(
x222 dx
Let x2 = t, then, d (x2) = dt ⇒ 2x dx = dt ⇒ dx = x2
dt
Ι = ∫ ++ 1tt
x2 .
x2
dt
= 2
1 ∫ ++ 1tt
12 dt
= 2
1 ∫
+
+
22
2
3
2
1t
1 dt
= 2
1 .
2
3
1 tan–1
+
2
3
2
1t
+ C
= 3
1 tan–1
+
3
1t2 + C =
3
1 tan–1
+
3
1x2 2
+ C.
Note: (i) ∫ [ f(x)]n f ′(x) dx = 1n
))x(f( 1n
+
+
(ii) ∫[ ]
′f x
f xn
( )
( ) dx =
n1
))x(f( n1
−
−
(iii) ∫ d x
x xn
( )+1 n ∈ N Take xn common & put 1 + x−n = t.
(iv) ∫( )
dx
x xnn
n21
1+−( ) n ∈ N, take xn common & put 1+x−n = tn
(v)
( )dx
x xn n
n
11
+∫ / take xn common as x and put 1 + x −n = t.
Self Practice Problems
1. ∫ +dx
xtan1
xsec2
Ans. �n |1 + tan x| + C
2. ∫ dxx
)nxsin(�Ans. – cos (�n x) + C
5 .5 .5 .5 . Integrat ion by Par t :Integrat ion by Par t :Integrat ion by Par t :Integrat ion by Par t :
(i) when you find integral ∫ dx)x(g then it will not contain arbitarary constant.
(ii) ∫ dx)x(g should be taken as same both terms.
(iii) the choice of f(x) and g(x) is decided by ILATE rule.
the function will come later is taken an integral function.
Ι → Inverse function
L → Logrithimic function
A → Algeberic function
T → Trigonometric function
E → Exponential function
Example : Evaluate : ∫− dxxtanx 1
Solution. ∫− dxxtanx 1
= (tan–1 x) 2
x2
– ∫ + 2x1
1 .
2
x2
dx
= 2
x2
tan–1 x – 2
1 ∫ +
−+
1x
11x2
2
dx = 2
x2
tan–1 x – 2
1 ∫ +
−1x
11
2 dx
= 2
x2
tan–1 x – 2
1 [x – tan–1 x] + C.
Example : Evaluate : ∫ + dx)x1log(x
Solution. ∫ + dx)x1log(x
= log (x + 1) . 2
x2
– ∫ +1x
1 .
2
x2
dx
= 2
x2
log (x + 1) – 2
1 ∫ +1x
x2
dx = 2
x2
log (x + 1) – 2
1 ∫ +
+−
1x
11x2
dx
= 2
x2
log (x + 1) – 2
1 ∫ +
−
1x
1x2
+ 1x
1
+ dx
= 2
x2
log (x + 1) – 2
1
++−∫ dx
1x
1)1x(
= 2
x2
log (x + 1) – 2
1
++− |1x|logx
2
x2
+ C
Example : Evaluate : ∫ x3sine x2 dx
Solution. Let Ι = ∫ x3sine x2dx. Then,
Ι = ∫ x3sine x2dx
⇒ Ι = e2x
−
3
x3cos – ∫
x2e2
−
3
x3cos dx
⇒ Ι = – 3
1 e2x cos 3x +
3
2 ∫ x3cose x2
dx
⇒ Ι = – 3
1 e2x cos 3x +
3
2
− ∫ dx
3
x3sine2
3
x3sine x2x2
⇒ Ι = – 3
1 e2x cos 3x +
9
2e2x sin 3x –
9
4 ∫ x3sine x2
dx
⇒ Ι = – 3
1 e2x cos 3x +
9
2e2x sin 3x –
9
4 Ι
⇒ Ι + 9
4 Ι =
9
e x2
(2 sin 3x – 3 cos 3x)
⇒9
13 Ι =
9
e x2
(2 sin 3x – 3 cos 3x)
⇒ Ι = 13
e x2
(2 sin 3x – 3 cos 3x) + C
Note :
(i) ∫ ex [f(x) + f ′(x)] dx = ex. f(x) + c
(ii) ∫ [f(x) + xf ′(x)] dx = x f(x) + c
Example : ∫xe 2)1x(
x
+ dx
Solution. ∫xe 2)1x(
11x
+
−+ dx
⇒ ∫ xe
+−
+ 2)1x(
1
)1x(
1 dx =
)1x(
ex
+ + c
Example : ∫ xe
−
−
xcos1
xsin1 dx
Solution. ∫xe
−
2
xsin2
2
xcos
2
xsin21
2 dx
⇒ ∫xe
−
2
xcoteccos
2
1 2 dx = – ex cot
2
x + c
Example : ∫
+
2)nx(
1)nx(n�
�� dx
Solution. put x = et
⇒ ∫te
+
2t
1nt� dt
⇒ ∫te
++−
2t
1
t
1
t
1nt� dt = et
−
t
1nt� + c
⇒ x
−
nx
1)nx(n�
�� + c
Self Practice Problems
1. ∫ dxxsinx Ans. – x cosx + sin x + C
2. ∫ dxex x2Ans. x2 ex – 2xex + 2ex + C
6.6 .6 .6 . Integration of Rational Algebraic Functions by using Partial Fractions:Integration of Rational Algebraic Functions by using Partial Fractions:Integration of Rational Algebraic Functions by using Partial Fractions:Integration of Rational Algebraic Functions by using Partial Fractions:
PARTIAL FRACTIONS :
If f(x) and g(x) are two polynomials, then )x(g
)x(f defines a rational algebraic function of a rational function
of x.
If degree of f(x) < degree of g(x), then )x(g
)x(f is called a proper rational function.
If degree of f(x) ≥ degree of g(x) then )x(g
)x(f is called an improper rational function
If )x(g
)x(f is an improper rational function, we divide f(x) by g(x) so that the rational function
)x(g
)x(f is
expressed in the form φ(x) + )x(g
)x(Ψ where φ(x) and ψ(x) are polynomials such that the degree of ψ(x) is
less than that of g(x). Thus, )x(g
)x(f is expressible as the sum of a polynomial and a proper rational
function.
Any proper rational function )x(g
)x(f 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
called the resolutions or decomposition of )x(g
)x(f into partial fractions.
The resolution of )x(g
)x(f 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 – a
2) .....(x – a
n). Then, we assume that
)x(g
)x(f =
1
1
ax
A
− +
2
2
ax
A
− + ..... +
n
n
ax
A
−
where A1, A
2, ...... A
n 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, a
Express px + q = A (differential co−efficient of denominator) + B.
Example : Evaluate : ∫++
+
1x4x
3x2
2 dx
Solution. ∫++
+
1x4x
3x2
2 dx
= ∫++
−+
1x4x
1)4x2(
2 dx
= ∫++
+
1x4x
4x2
2 dx – ∫++ 1x4x
1
2 dx
= ∫ t
dt –
( )∫
−+22 3)2x(
1 dx, where t = x2 + 4x + 1
= 2 t – log | (x + 2) + 1x4x2 ++ | + C
= 2 1x4x2 ++ – log | x + 2 + 1x4x2 ++ | + C
Example : Evaluate : ∫ +− xx)5x( 2 dx
Solution. Let (x – 5) = λ . dx
d (x2 + x) + µ. Then,
x – 5 = λ (2x + 1) + µ.
Comparing coefficients of like powers of x, we get
1 = 2λ and λ + µ = – 5 ⇒ λ = 2
1 and µ = –
2
11
∫ +− xx)5x( 2 dx
= ∫
−+
2
11)1x2(
2
1 xx2 + dx
= ∫ + )1x2(2
1 xx2 + dx –
2
11 ∫ + xx2
dx
= 2
1∫ + )1x2( xx2 + dx –
2
11 ∫ + xx2
dx
= 2
1
∫ t dt – 2
11 ∫
−
+
22
2
1
2
1x dx where t = x2 + x
= 2
1 .
2/3
t 2/3
– 2
11
−
+
+
22
2
1
2
1x
2
1x
2
1
– 2
1 .
2
2
1
log
−
++
+
22
2
1
2
1x
2
1x + C
= 3
1 t3/2 –
2
11
++
+−+
+xx
2
1xn
8
1xx
4
1x2 22� + C
= 3
1(x2 + x)3/2 –
2
11
++
+−+
+xx
2
1xn
8
1xx
4
1x2 22� + C
Self Practice Problems
1. ∫ ++
+
3xx
1x2 dx Ans.
2
1 log |x2 + x + 3| +
11
1 tan–1
+
11
1x2 + C
2. ∫+−
−
1x5x3
5x6
2 dx Ans. 2 1x5x3 2 +− + C
3. ∫ ++− 2xx1)1x( dx
Ans.3
1 (x2 + x + 1)3/2 –
8
3 (2x + 1) 2xx1 ++ –
16
9 log (2x +1 + 2 1xx2 ++ ) + C
9.9 .9 .9 . Integrat ion of trigonometric functionsIntegrat ion of trigonometric functionsIntegrat ion of trigonometric functionsIntegrat ion of trigonometric functions
(i) ∫ xxxxsinsinsinsinbbbbaaaa xxxxdddd 2222+ OR ∫ xxxxcoscoscoscosbbbbaaaa xxxxdddd 2222+
OR ∫ xxxxcoscoscoscosccccxxxxcoscoscoscosxxxxsinsinsinsinbbbbxxxxsinsinsinsinaaaa xxxxdddd 22222222 ++
Multiply Nr & Dr by sec² x & put tan x = t.
(ii) ∫ sinxsinxsinxsinxbbbbaaaa xxxxdddd+
OR ∫ cosxcosxcosxcosxbbbbaaaa xxxxdddd+
OR ∫ xxxxcoscoscoscosccccxxxxsinsinsinsinbbbbaaaa xxxxdddd++
Hint:
Convert sines & cosines into their respective tangents of half the angles and then,
⇒ 3 sin x + 2 cos x = λ (–3 sin x + 2 cos x) + µ (3 cos x + 2 sin x )
Comparing the coefficients of sin x and cos x on both sides, we get
– 3λ + 2µ = 3 and 2λ + 3µ = 2 ⇒ µ = 13
12 and λ = –
13
5
∴ Ι = ∫ +
+λ++−µ
xsin2xcos3
)xsin2xcos3()xcos2xsin3( dx
= λ ∫ dx.1 + µ ∫ +
+−
xsin2xcos3
xcos2xsin3 dx
= λ x + µ ∫ t
dt, where t = 3 cos x + 2 sin x
= λ x + µ �n | t | + C = 13
5− x +
13
12 �n | 3 cos x + 2 sin x | + C
Example : Evaluate : ∫ ++
+
3xcos2xsin
2xcos3 dx
Solution. We have,
Ι = ∫ ++
+
3xcos2xsin
2xcos3 dx
Let 3 cos x + 2 = λ (sin x + 2 cos x + 3) + µ (cos x – 2 sin x) + ν
Comparing the coefficients of sin x, cos x and constant term on both sides, we get
λ – 2µ = 0, 2λ + µ = 3, 3λ + ν = 2
⇒ λ = 5
6, µ
5
3 and ν = –
5
8
∴ Ι = ∫ ++
ν+−µ+++λ
3xcos2xsin
)xsin2x(cos)3xcos2x(sin dx
⇒ Ι = λ ∫ µ+dx ∫ ++
−
3xcos2xsin
xsin2xcos dx + ν ∫ ++ 3xcos2xsin
1 dx
⇒ Ι = λ x + µ log | sin x + 2 cos x + 3 | + ν Ι1, where
Ι1 = ∫ ++ 3xcos2xsin
1 dx
Putting, sin x = 2/xtan1
2/xtan22+
, cos x = 2/xtan1
2/xtan12
2
+
− we get
Ι1 = ∫
++
−+
+3
2/xtan1
)2/xtan1(2
2/xtan1
2/xtan2
1
2
2
2
dx
= ∫ ++−+
+
)2/xtan1(32/xtan222/xtan2
2/xtan122
2
dx
= ∫ ++ 52/xtan22/xtan
2/xsec2
2
dx
Putting tan 2
x = t and
2
1 sec2
2
x = dt or sec2
2
x dx = 2 dt, we get
Ι1 = ∫ ++ 5t2t
dt22
= 2 ∫ ++ 22 2)1t(
dt =
2
2 tan–1
+
2
1t = tan–1
+
2
12
xtan
Hence, Ι = λx + µ log | sin x + 2 cos x + 3 | + ν tan–1
+
2
12
xtan
+ C
where λ = 5
6, µ =
5
3 and ν = –
5
8
Example : ∫ + xcos31
dx2
Solution. = ∫ + 4xtan
dxxsec2
2
= 2
1 tan–1
2
xtan + C
Self Practice Problems
1. ∫ +
+
xcos4xsin5
xcos5xsin4 dx Ans.
41
40x +
41
9 log |5sinx + 4cosx| + C
10 .10 .10 .10 . Integration of type Integration of type Integration of type Integration of type ∫ dxdxdxdxxxxxcoscoscoscosx.x.x.x.sinsinsinsin nnnnmmmmCase - ΙΙΙΙ
If m and n are even natural number then converts higher power into higher angles.
Case - ΙΙΙΙΙΙΙΙ
If at least m or n is odd natural number then if m is odd put cosx = t and vice-versa.
Case - ΙΙΙ ΙΙΙ ΙΙΙ ΙΙΙ
When m + n is a negative even integer then put tan x = t.
Example: ∫ dxxcosxsin 45
Solution. put cos x = t ⇒ – sinx dx = dt
= – ∫ − 22 )t1( . t4 . dt
= – ∫ +− )1t2t( 24t4 dt
= – ∫ +− )tt2t( 468 dt
= – 9
t9
+ 7
t2 7
– 5
t5
+ c
= – 9
xcos9
+ 2 7
xcos7
– 5
xcos5
+ c Ans.
Example : ∫− dx)x(cos)x(sin 3/73/1
Solution. ∫− dx)x(cos)x(sin 3/73/1
= ∫3/1)x(tan
xcos
12 dx
put tanx = t ⇒ sec2x dx = dt
= dtt 3/1
∫
= 4
3 t4/3 + c
= 4
3 (tanx)4/3 + c Ans.
Example : ∫ dxxcosxsin 42
Solution.8
1 ∫ + dx)x2cos1(x2sin2
= 8
1 ∫ dxx2sin2
+ 8
1 ∫ dxx2cosx2sin2
= 16
1 ∫ +−
16
1dx)x4cos1(
3
x2sin3
= 16
1 –
64
x4sin +
48
x2sin3
+ c
11 .11 .11 .11 . Integration of typeIntegration of typeIntegration of typeIntegration of type
∫ 1111xxxxKKKKxxxx 1111xxxx 22224444 2222++
± dx where K is any constant.
Divide Nr & Dr by x² & put x ∓ x
1 = t.
Example : ∫ 42
2
xx1
x1
++
−dx
Solution. ∫ 1x
1x
dxx
11
2
2
2
++
−
x + x
1 = t
⇒ – ∫ 1t
dt2 −
– 2
1 �n
1t
1t
+
− + C
– 2
1 �n
1x
1x
1x
1x
++
−+
+ C
Example : Evaluate : ∫ + 1x
14 dx
Solution. We have,
Ι = ∫ + 1x
14 dx
⇒ Ι = ∫+
2
2
2
x
1x
x
1
dx
⇒ Ι = 2
1 ∫
+2
2
2
x
1x
x
2
dx
⇒ Ι = 2
1 ∫
+
−
−
+
+
2
2
2
2
2
2
x
1x
x
11
x
1x
x
11
dx
⇒ Ι = 2
1 ∫
+
+
2
2
2
x
1x
x
11
dx – 2
1 ∫
+
−
2
2
2
x
1x
x
11
dx
⇒ Ι = 2
1 ∫
+
−
+
2x
1x
x
11
2
2
dx – 2
1 ∫
−
+
−
2x
1x
x
11
2
2
dx
Putting x – x
1 = u in 1st integral and x +
x
1 = ν in 2nd integral, we get
Ι = 2
1
( )∫+
22 2u
du
– 2
1 ( )∫
−ν
ν22 2
d
= 22
1 tan–1
2
u –
2
1
22
1 log
2
2
+ν
−ν + C
= 22
1 tan–1
−
2
x/1x –
24
1 log
2x/1x
2x/1x
++
−+ + C
= 22
1 tan–1
−
x2
1x2
– 24
1 log
12xx
1x2x2
2
++
+− + C
Self Practice Problem :
1. ∫ 1x7x
1x24
2
+−
− dx Ans.
6
1 �n
3x
1x
3x
1x
++
−+
+ C
2. ∫ xtan dx Ans.2
1 tan–1
2
y +
22
1 �n
2y
2y
+
− + C where y = tan x –
xtan
1
1 2 .1 2 .1 2 .1 2 . Integrat ion of typeIntegrat ion of typeIntegrat ion of typeIntegrat ion of type
∫ ++ qqqqpxpxpxpxb)b)b)b)(ax(ax(ax(ax dxdxdxdx OR ( )∫ +++ qqqqpxpxpxpxccccbxbxbxbxaxaxaxax dxdxdxdx2222 ; put px + q = t2.
Example: Evaluate : ∫ +− 1x)3x(
1 dx
Solution. Let Ι = ∫ +− 1x)3x(
1 dx
Here, P and Q both are linear, so we put Q = t2 i.e. x + 1 = t2 and dx = 2t dt
∴ Ι = 22
t
t2
)31t(
1∫ −−
dt
⇒ Ι = 2 ∫ − 22 2t
dt = 2 .
)2(2
1 log
2t
2t
+
− + C
⇒ Ι = 2
1 log
21x
21x
++
−+ + C.
Example : Evaluate : ∫ +++
+
1x)3x3x(
2x2 dx
Solution. Let Ι = ∫ +++
+
1x)3x3x(
2x2 dx
Putting x + 1 = t2, and dx = 2t dt, we get Ι = ∫+−+−
+
2222
2
t}3)1t(3)1t{(
dtt2)1t(
⇒ Ι = 2 ∫ ++
+
1tt
)1t(24
2
dt = 2 ∫++
+
1t
1t
t
11
2
2
2
dt
⇒ Ι = 2 ( )∫ ′+
22 3u
du
where t – t
1 = u.
⇒ Ι = 3
2 tan–1
3
u + C =
3
2 tan–1
−
3
t
1t
+ C
⇒ Ι = 3
2 tan–1
−
3t
1t2
+ C = 3
2 tan–1
+ )1x(3
x + C
13 .13 .13 .13 . Integration of typeIntegration of typeIntegration of typeIntegration of type
∫+++ rrrrqxqxqxqxpxpxpxpxb)b)b)b)(ax(ax(ax(ax dxdxdxdx 2222 , put ax + b = tttt1111 ; ∫
1 4 .1 4 .1 4 .1 4 . Integrat ion of typeIntegrat ion of typeIntegrat ion of typeIntegrat ion of type
∫ −
− xxxxββββ ααααxxxx dx or ( )( )∫ −− xxxxββββααααxxxx ; put x = αααα cos2 θθθθ + ββββ sin2 θθθθ
∫ −
−ββββxxxx ααααxxxx dx or ( )( )∫ −− ββββxxxxααααxxxx ; put x = αααα sec2 θ −θ −θ −θ − ββββ tan2 θθθθ
( )( )∫ −− ββββxxxxααααxxxx dxdxdxdx; put x −−−− αααα = t2 or x −−−− ββββ = t2.
1 5 .1 5 .1 5 .1 5 . Reduct ion formula of Reduct ion formula of Reduct ion formula of Reduct ion formula of ∫ dxxntan , , , , ∫ dxxncot , , , , ∫ dxxnsec ,,,,
Clearly, Sn is area very close to the area of the region bounded by curve y = f(x), x–axis and the ordinates
x = a, x = b.
Hence ∫b
a
dx)x(f = ∞→n
Lt Sn
∫b
a
dx)x(f = ∞→n
Lt ∑−
=
+
1n
0r
)rha(fh = ∞→n
Lt ∑−
=
−1n
0rn
ab f
−+
n
r)ab(a
Note :
1. We can also write
Sn = hf(a + h) + hf (a + 2h) + .........+ hf(a + nh) and ∫
b
a
dx)x(f = ∞→n
Lt ∑=
−n
1rn
ab f
−+ r
n
aba
2. If a = 0, b = 1, ∫1
0
dx)x(f = ∞→nLt ∑
−
=
1n
0r n
rf
n
1
Steps to express the limit of sum as definte integral
Step 1. Replace n
r by x,
n
1 by dx and ∞→n
Lt ∑ by ∫
Step 2. Evaluate ∞→nLt
n
r by putting least and greatest values of r as lower and upper limits respectively..
For example ∞→nLt ∑
=
pn
1rn
rf
n
1 = ∫
p
0
dx)x(f (∵ 1r
n n
rLt
=∞→
= 0,
nprn n
rLt
=∞→
= p)
Illustration 25 : Evaluate
∞→nLt
++
++
++
+ n2
1.........
n3
1
n2
1
n1
1
Sol. ∞→nLt
++
++
++
+ n2
1.........
n3
1
n2
1
n1
1
= ∞→nLt ∑
=+
n
1rnr
1
= ∞→nLt ∑
=
n
1rn
1
1n
r
1
+
= ∫ +
1
01x
dx = [ ]10e )1x(log + = log
e2.
Illustration 26 : Evaluate ∞→nLt
++
+
++
+
++
+
+
n5
3.........
3n
3n
2n
2n
1n
1n222222
Sol. ∞→nLt ∑
= +
+n2
1r22 rn
rn = ∞→n
Lt ∑=
n2
1rn
1 2
n
r1
n
r1
+
+
∵ ∞→nLt
n
r = 0, when r = 1, lower limit = 0
and ∞→nLt
n
r = ∞→n
Lt
n
n2 = 2, when r = 2n, upper limit = 2
∫ +
+2
0
2x1
x1 dx = ∫ +
2
0
2x1
1 dx +
2
1 ∫ +
2
0
2x1
x2 dx
= tan–1x]20 +
2
0
2e )x1(log
2
1
+
= tan–1 2 + 2
1 log
e5
Illustration 27 : Evaluate
∞→nLt
n
1
nn
!n
Sol. Let y = ∞→nLt
n
1
nn
!n
loge y = ∞→n
Lt n
1 log
e
nn
!n
= ∞→nLt
n
1log
e
nn
n........3.2.1
= ∞→nLt
n
1
++
+
+
n
nlog.....
n
3log
n
2log
n
1log eeee
= ∞→nLt
n
1 ∑=
n
1r
en
rlog
=
1
0
e
1
0
e xxlogxdxxlog
−=∫
= (0 – 1) – +→0x
Lt x logex + 0
= – 1 – 0 = –1
⇒ y = e
1
Self Practice Problems :
Evaluate the following limits
1. ∞→nLt
+++
++
++
22222 nn
1....
n2n
1
nn
1
n
1Ans. 2 ( )12 −
2. ∞→nLt
++
++
++
+ n5
1.......
n3
1
n2
1
n1
1Ans. log
e5
3. ∞→nLt 2n
1
π++
π+
π+
π
n4
nsinn........
n4
3sin3
n4
2sin2
n4sin 3333
Ans. 29
2
π (52 – 15π)
4. ∞→nLt ∑
−
=
1n
0r22 rn
1
−Ans.
2
π
5. ∞→nLt
n
3
−+++
++
++
++
)1n(3n
n......
9n
n
6n
n
3n
n1 Ans. 2
PART – J
Reduction Formulae in Definite Integrals
1. If Ιn = ∫
π
2
0
n xsin dx , then show that Ιn =
−
n
1n Ι
n–2
Proof : Ιn = ∫
π
2
0
n xsin dx
Ιn = [ ]201n xcosxsin
π
−− + ∫
π
−−
2
0
22n dxxcos.xsin)1n(
= (n – 1) ∫
π
− −
2
0
22n dx)xsin1(.xsin
= (n – 1) ∫
π
− −−
2
0
2n )1n(dxxsin ∫
π
2
0
n dxxsin
Ιn + (n – 1) Ι
n = (n – 1) Ι
n–2
Ιn =
−
n
1n Ι
n–2
Note : 1. ∫
π
2
0
n dxxsin = ∫
π
2
0
n dxxcos
2. Ιn =
−
n
1n
−
−
2n
3n
−
−
4n
5n ..... Ι
0 or Ι
1
according as n is even or odd. Ι0 =
2
π, Ι
1 = 1
Hence Ιn =
−
−
−
−
−
π
−
−
−
−
−
oddisnif1.3
2........
4n
5n
2n
3n
n
1n
evenisnif2
.2
1........
4n
5n
2n
3n
n
1n
2. If Ιn = ∫
π
4
0
n dxxtan , then show that Ιn + Ι
n–2 =
1n
1
−
Sol. Ιn = ∫
π
−4
0
2n)x(tan . tan2x dx
= ∫
π
−4
0
2n)x(tan (sec2x – 1) dx
= ∫
π
−4
0
2n)x(tan sec2x dx – ∫
π
−4
0
2n)x(tan dx
= 4
0
1n
1n
)x(tan
π−
− – Ι
n–2
Ιn =
1n
1
− – Ι
n–2
∴ Ιn + Ι
n–2 =
1n
1
−
3. If Ιm,n
= ∫
π
2
0
nm dxxcos.xsin , then show that Ιm,n
= nm
1m
+
− Ι
m–2 , n
Sol. Ιm,n
= ∫
π
−2
0
n1m )xcosx(sinxsin dx
= 2
0
1n1m
1n
xcos.xsin
π+−
+− + ∫
π
+
+
2
0
1n
1n
xcos (m – 1) sinm–2 x cos x dx
=
+
−
1n
1m ∫
π
−2
0
2n2m xcos.xcos.xsin dx
=
+
−
1n
1m ( )∫
π
− −
2
0
nmn2m xcos.xsinxcos.xsin dx
=
+
−
1n
1m Ι
m–2,n –
+
−
1n
1m Ι
m,n
⇒
+
−+
1n
1m1 Ι
m,n =
+
−
1n
1m Ι
m–2,n
Ιm,n
=
+
−
nm
1m Ι
m–2,n
Note : 1. Ιm,n
=
+
−
nm
1m
−+
−
2nm
3m
−+
−
4nm
5m ........ Ι
0,n or Ι
1,n according as m is even or odd.
Ι0,n
= ∫
π
2
0
n dxxcos and Ι1,n
= ∫
π
2
0
n dxxcos.xsin = 1n
1
+
2. Walli’s Formula
Ιm,n
=
−+−++
−−−−−−
π
−+−++
−−−−−−
otherwise)........4nm()2nm()nm(
).......5n()3n()1n.........()5m()3m()1m(
evenaren,mbothwhen2)........4nm()2nm()nm(
).......5n()3n()1n.........()5m()3n()1n(
Illustration 28 : Evaluate ∫
π
π−
+2
2
22 dx)xcosx(sinxcosxsin
Sol. Given integral = ∫
π
π−
2
2
23 dxxcosxsin + ∫
π
π−
2
2
32 dxxcosxsin
= 0 + 2 ∫
π
2
0
32 dxxcosxsin (∵ sin3x cos2x is odd and sin2x cos3x is even)
= 2. 1.3.5
2.1 =
15
4
Illustration 29 : Evaluate ∫π
0
65 dxxcosxsinx
Sol. Let Ι = ∫π
0
65 dxxcosxsinx
Ι = ∫π
−π−π−π
0
65 dx)x(cos)x(sin)x(
= π ∫π
0
65 dxxcos.sin – ∫π
0
65 dxxcos.xsinx
⇒ 2Ι = π . 2 ∫
π
2
0
65 dxxcos.xsin
Ι = π 1.3.5.7.9.11
1.3.5.2.4
Ι = 693
8π
Illustration 30 : Evaluate ∫ −
1
0
53 dx)x1(x
Sol. Put x = sin2θ ⇒ dx = 2 sin θ cos θ dθL . L : x = 0 ⇒ θ = 0
U.L. : x = 1 ⇒ θ = 2
π
∴ ∫ −
1
0
53 dx)x1(x = ∫
π
θθ
2
0
526 )(cossin 2 . sin θ . cos θ dθ
= 2 . ∫
π
θθθ
2
0
117 dcossin
= 2 . 2.4.6.8.10.12.14.16.18
2.4.6.8.10.2.4.6=
504
1
Self Practice Problems:
Evaluate the following
1. ∫
π
2
0
5 dxxsin Ans.15
18
2. ∫
π
2
0
45 dxxcosxsin Ans.315
8
3. ∫ −
1
0
16 dxxsinx Ans.14
π –
245
16
4. ( )∫ −
a
0
2
722 dxxax Ans.
9
a9
5. ∫ −
2
0
2/3 dxx2x Ans.2
π
KEY CONCEPTS1. DEFINITION :
If f & g are functions of x such that g′(x) = f(x) then the function g is called a PRIMITIVE OR
ANTIDERIVATIVE OR INTEGRAL of f(x) w.r.t. x and is written symbolically as
∫ f(x) dx = g(x) + c ⇔ d
dx{g(x) + c} = f(x), where c is called the constant of integration.
2. STANDARD RESULTS :
(i) ∫ (ax + b)n dx = ( )
( )ax b
a n
n++
+1
1 + c n ≠ −1 (ii) ∫ dx
ax b+ =
1
a ln (ax + b) + c
(iii) ∫ eax+b dx = 1
a eax+b + c (iv) ∫ apx+q dx =
1
p
a
na
px q+
� (a > 0) + c
(v) ∫ sin (ax + b) dx = − 1
a cos (ax + b) + c (vi) ∫ cos (ax + b) dx =
1
a sin (ax + b) + c
(vii) ∫ tan(ax + b) dx = 1
a
ln sec (ax + b) + c (viii) ∫ cot(ax + b) dx = 1
a ln sin(ax + b)+ c
(ix) ∫ sec² (ax + b) dx = 1
a tan(ax + b) + c (x) ∫ cosec²(ax + b) dx = −
1
acot(ax + b)+ c
(xi) ∫ sec (ax + b) . tan (ax + b) dx = a
1 sec (ax + b) + c
(xii) ∫ cosec (ax + b) . cot (ax + b) dx = a
1− cosec (ax + b) + c
(xiii) ∫ secx dx = ln (secx + tanx) + c OR ln tan π4 2
+
x+ c
(xiv) ∫ cosec x dx = ln (cosecx − cotx) + c OR ln tan 2
x + c OR − ln (cosecx + cotx)
(xv) ∫ sinh x dx = cosh x + c (xvi) ∫ cosh x dx = sinh x + c (xvii) ∫ sech²x dx = tanh x + c
(xviii) ∫ cosech²x dx = − coth x + c (xix) ∫ sech x . tanh x dx = − sech x + c
(xx) ∫ cosech x . coth x dx = − cosech x + c (xxi) ∫22 xa
xd
− = sin−1
a
x + c
(xxii) ∫ 22 xa
xd
+ =
a
1 tan−1
a
x + c (xxiii) ∫ 22 axx
xd
− =
a
1 sec−1
a
x + c
(xxiv) ∫ d x
x a2 2+ = ln
++ 22 axx OR sinh−1
a
x + c
(xxv) ∫22 ax
xd
− = ln
−+ 22 axx OR cosh−1
a
x + c
(xxvi) ∫ 22 xa
xd
− =
a2
1 ln
xa
xa
−+
+ c (xxvii) ∫ 22ax
xd
− =
a2
1 ln
ax
ax
+−
+ c
(xxviii) ∫ 22xa − dx =
2
x 22
xa − + 2
a2
sin−1
a
x + c
(xxix) ∫ 22 ax + dx = 2
x 22 ax + +
2
a2
sinh−1
a
x + c
(xxx) ∫ 22 ax − dx = 2
x 22 ax − −
2
a2
cosh−1
a
x + c
(xxxi) ∫ eax. sin bx dx = 22
ax
ba
e
+ (a sin bx − b cos bx) + c
(xxxii) ∫ eax . cos bx dx = 22
ax
ba
e
+ (a cos bx + b sin bx) + c
3. TECHNIQUES OF INTEGRATION :(i) Substitution or change of independent variable .
Integral I = ∫ f(x) dx is changed to ∫ f(φ (t)) f ′ (t) dt , by a suitable substitution
x = φ (t) provided the later integral is easier to integrate .
(ii) Integration by part : ∫ u.v dx = u ∫ v dx − ∫
∫ xdv.
xd
ud dx where u & v are differentiable
function . Note : While using integration by parts, choose u & v such that
(a) ∫ v dx is simple & (b) ∫
∫ xdv.
xd
ud dx is simple to integrate.
This is generally obtained, by keeping the order of u & v as per the order of the letters in ILATE,where ; I − Inverse function, L − Logarithmic function ,A − Algebraic function, T − Trigonometric function & E − Exponential function
(iii) Partial fraction , spiliting a bigger fraction into smaller fraction by known methods .
4. INTEGRALS OF THE TYPE :
(i) ∫ [ f(x)]n f ′(x) dx OR ∫ [ ]n)x(f
)x(f ′ dx put f(x) = t & proceed .
(ii)dx
ax bx c2 + +∫ , dx
ax bx c2 + +∫ , ax bx c2 + +∫ dx
Express ax2 + bx + c in the form of perfect square & then apply the standard results .
(iii)px q
ax bx c
++ +∫ 2 dx ,
px q
ax bx c
+
+ +∫ 2
dx .
Express px + q = A (differential co-efficient of denominator) + B .
(iv) ∫ ex [f(x) + f ′(x)] dx = ex . f(x) + c (v) ∫ [f(x) + xf ′(x)] dx = x f(x) + c
(vi) ∫)1x(x
xdn +
n ∈ N Take xn common & put 1 + x−n = t .
(vii) ∫( ) n
)1n(n2 1xx
xd−
+ n ∈ N , take xn common & put 1+x−n = tn
(viii)( )
dx
x xn nn
11
+∫ /
take xn common as x and put 1 + x −n = t .
(ix) ∫xsinba
xd2+
OR ∫xcosba
xd2+
OR ∫xcoscxcosxsinbxsina
xd22 ++
Multiply ..rN & ..rD by sec² x & put tan x = t .
(x) ∫xsinba
xd
+ OR ∫
xcosba
xd
+ OR ∫
xcoscxsinba
xd
++
Hint :Convert sines & cosines into their respective tangents of half the angles , put tan 2
x = t
(xi) ∫nxsin.mxcos.
cxsin.bxcos.a
++++
� dx . Express Nr ≡ A(Dr) + B
xd
d (Dr) + c & proceed .
(xii) ∫1xKx
1x24
2
+++
dx OR ∫1xKx
1x24
2
++−
dx where K is any constant .
Hint : Divide Nr & Dr by x² & proceed .
(xiii)dx
ax b px q( )+ +∫ & ( )dx
ax bx c px q2 + + +∫ ; put px + q = t2 .
(xiv) dx
ax b px qx r( )+ + +∫ 2
, put ax + b = 1
t ;
( )dx
ax bx c px qx r2 2+ + + +∫ , put x =
1
t
(xv)x
x
−−∫
αβ
dx or ( ) ( )x x− −∫ α β ; put x = α cos2 θ + β sin2 θ
x
x
−−∫
αβ
dx or ( ) ( )x x− −∫ α β ; put x = α sec2 θ − β tan2 θ
( ) ( )dx
x x− −∫
α β ; put x − α = t2 or x − β = t2 .
DEFINITE INTEGRAL
1. ∫b
a
f(x) dx = F(b) − F(a) where ∫ f(x) dx = F(x) + c
VERY IMPORTANT NOTE : If ∫b
a
f(x) dx = 0 ⇒ then the equation f(x) = 0 has atleast one
root lying in (a , b) provided f is a continuous function in (a , b) .
2. PROPERTIES OF DEFINITE INTEGRAL :
P−−−−1 ∫b
a
f(x) dx = ∫b
a
f(t) dt provided f is same P −−−− 2 ∫b
a
f(x) dx = − ∫a
b
f(x) dx
P−−−−3 ∫b
a
f(x) dx = ∫c
a
f(x) dx + ∫b
c
f(x) dx , where c may lie inside or outside the interval [a, b] . This property
to be used when f is piecewise continuous in (a, b) .
P−−−−4 ∫−
a
af(x) dx = 0 if f(x) is an odd function i.e. f(x) = − f(−x) .
= 2 ∫a
0
f(x) dx if f(x) is an even function i.e. f(x) = f(−x) .
P−−−−5 ∫b
a
f(x) dx = ∫b
a
f(a + b − x) dx , In particular ∫a
0
f(x) dx = ∫a
0
f(a − x)dx
P−−−−6 ∫a2
0
f(x) dx = ∫a
0
f(x) dx + ∫a
0
f(2a − x) dx = 2 ∫a
0
f(x) dx if f(2a − x) = f(x)
= 0 if f(2a − x) = − f(x)
P−−−−7 ∫an
0
f(x) dx = n ∫a
0
f(x) dx ; where‘a’is the period of the function i.e. f(a + x) = f(x)
P−−−−8a nT
b nT
+
+
∫ f(x) dx = a
b
∫ f(x) dx where f(x) is periodic with period T & n ∈ I .
P−−−−9ma
na
∫ f(x) dx = (n − m) 0
a
∫ f(x) dx if f(x) is periodic with period 'a' .
P−−−−10 If f(x) ≤ φ(x) for a ≤ x ≤ b then ∫b
a
f(x) dx ≤ ∫b
a
φ (x) dx
P−−−−11 xd)x(fb
a
∫ ≤ ∫b
a
f(x)dx .
P−−−−12 If f(x) ≥ 0 on the interval [a, b] , then a
b
∫ f(x) dx ≥ 0.
3. WALLI’S FORMULA :
∫π 2/
0
sinnx . cosmx dx = [ ][ ]
2or1)....4nm()2nm()nm(
2or1)....3m()1m(2or1)....5n()3n()1n(
−+−++−−−−−
K
Where K = 2
π if both m and n are even (m, n ∈ N) ;
= 1 otherwise4. DERIVATIVE OF ANTIDERIVATIVE FUNCTION :
If h(x) & g(x) are differentiable functions of x then ,
xd
d ∫)x(h
)x(g
f(t) dt = f [h (x)] . h′(x) − f [g (x)] . g′(x)
5. DEFINITE INTEGRAL AS LIMIT OF A SUM :
∫b
a
f(x) dx = ∞→nLimit h [f (a) + f (a + h) + f (a + 2h) + ..... + f ( )a n h+ −1 ]
= 0hLimit
→ h ∑=
−
r
n
0
1
f (a + rh) where b − a = nh
If a = 0 & b = 1 then , ∞→nLimit h ∑
=
−
r
n
0
1
f (rh) = ∫1
0
f(x) dx ; where nh = 1 OR
∞→nLimit
n
1
1n
1r
−
=∑ f
n
r = ∫
1
0
f(x) dx .
6. ESTIMATION OF DEFINITE INTEGRAL :
(i) For a monotonic decreasing function in (a , b) ; f(b).(b − a) < ∫b
a
f(x) dx < f(a).(b − a) &
(ii) For a monotonic increasing function in (a , b) ; f(a).(b − a) < ∫b
2 f ′ (x) w.r.t. x4 , where f (x) = tan −1x + ln 1+ x − ln 1− x
Q.21 ∫ ++
)1x(x
dx)1x(3 Q.22
dx
x xsin cos2 2
3∫ Q.23 ∫ ++
+dx
)1xe(
xx2x
2
Q.24 ∫ xsec21
xsec.
xcotxeccos
xcotxeccos
++−
dx Q.25 ∫ x2sin97
xsinxcos
−−
dx Q.26 ∫xeccosxsec
xd
+ dx
Q.27 ∫ xsecxsin
xd
+ Q.28 ∫ tan x.tan 2x.tan 3x dx Q.29 ( )
dx
x xsin sin 2 +∫
α
Q.30 dx)xcosxsinx)(xsinxcosx(
x2
∫ +−Q.31 ∫ xcosxsin23
xcos2xsin43
++++
dx
Q.32 ∫( )�n x x
x
cos cos
sin
+ 2
2dx
Q.33 sin
sin cos
x
x x+∫ dx Q.34 ∫ xtanxsin
xd
+Q.35 ∫ −
+dx
)1x(
1x332
2
Q.36 ∫+
dxxsin
)xcosxsinx(e2
3xcos
Q.37 ∫( )
( )
ax b dx
x c x ax b
2
2 2 2 2
−
− +Q.38 ∫
( )2
2x
x1)x1(
x2e
−−
−dx
Q.39 ∫ 2/3)x10x7(
x
2−− dx Q.40 ∫ ( ) 2/32 1x
xlnx
− dx Q.41 ∫
1
13
−+
x
x
dx
x
Q.42 2 3
2 3
1
1
−+
+−∫
x
x
x
x dx Q.43
cot tan
sin
x x
x
−+∫
1 3 2 dx Q.44 ∫ +
−+−+−dx
)1x(x
7x4x2x8x7x4222
2345
Q.45 ∫ ( ) 1x
xd
3x3x
2x2 +++
+Q.46 ∫
2 2
2
− −x x
x dx Q.47 ∫ )x()x()x(
xd
β−α−α−
Q.48 ∫++−+ 1x4x6x4x
dxx
234Q.49 ∫ xsin
x2cos dx Q.50 ∫ +α−
+42
2
xcosx21
dx)x1(α∈(0, π)
EXERCISE–2
Q.1 ∫π
0
x dx
x x9 2 2cos sin+ Q.2 1 2
1 20
2 −+z sin
sin
x
xdx
π
Q.3 Evaluate In = ∫
e
1
(lnn x) dx hence find I3.
Q.4 ∫π 2/
0
sin2x · arc tan(sinx) dx Q.5 ∫π 2/
0
cos4 3x . sin2 6x dx Q.6 ∫π 4/
0
x dx
x x xcos (cos sin )+
Q.7 Let h (x) = (fog) (x) + K where K is any constant. If ( ))x(hdx
d = –
)x(coscos
xsin2
then compute the
value of j (0) where j (x) = ∫)x(f
)x(g
dt)t(g
)t(f, where f and g are trigonometric functions.
Q.8 Find the value of the definite integral ∫π
+0
dxxcos2xsin2 .
Q.9 Evaluate the integral : x x x x+ − + − −
∫ 2 2 4 2 2 4
3
5
dx
Q.10 If P = ∫∞
0
x
xdx
2
41+; Q = ∫
∞
0
xdx
x1 4+ and R = ∫
∞
0
dx
x1 4+ then prove that
(a) Q = π4
, (b) P = R, (c) P – 2 Q + R = π
2 2
Q.11 Prove that x n x n a b x nab
x a x bdx
b a
a b
n
a
b n n− − −− + − + +
+ +=
−+z
1 2
2 2
1 12 1
2
b ge j( )( )
( ) ( ) ( )
Q.12 x x
xdx
4 4
2
0
11
1
( )−+∫ Q.13 ∫
1
02
2
x1
xln.x
− dx Q.14 Evaluate: ∫
− +
−2
22
2
dx4x
xx
Q.15 ∫ +−
3
0
2
1 dxx1
x2sin Q.16
0
2π/
∫ ( )xsin
xcosbxsina
4+
+π dx Q.17
0
2π
∫dx
x2 2+ sin
Q.18 ∫π
+π2
0
x dx2
x
4cose Q.19 ∫
−
2
2 2x
1xx12x7x10x3x22
23567
+++−−−+
dx
Q.20 Let α, β be the distinct positive roots of the equation tan x = 2x then evaluate ∫ βα1
0
dx)xsin·x(sin ,
independent of α and β.
Q.21 ∫π 4/
0 x2sin10
xsinxcos
+−
dx Q.22 ∫π
0
( )sec tan
tan
ax b x x
x
++4 2
dx (a,b>0) Q.23 Evaluate: ∫π
++
02
dx)xcos1(
xsin)3x2(
Q.24 If a1, a
2 and a
3 are the three values of a which satisfy the equation
∫ +1
0
3dx)xcosax(sin – ∫−π
1
0
dxxcosx2
a4 = 2
then find the value of 1000( 23
22
21 aaa ++ ).
Q.25 Show that 0
p q+
∫π
| cos x| dx = 2q + sinp where q ε N & − < <π π2 2
p
Q.26 Show that the sum of the two integrals ∫−
−
5
4
e(x+5)² dx + 3 ∫3/2
3/1
e9(x− 2/3)² dx is zero.
Q.27 ∫ +−
−1
02
1
dx1xx
xsinQ.28 ∫
π 2/
0xcosbxsina
xsin2222
2
+dx (a>0, b>0) Q.29 �n
x
x
x
x
1
1 11
1 3
2
+− −−
∫ dx
Q.30 0
2π/
∫ tan−1
−−+
−++
xsin1xsin1
xsin1xsin1dx Q.31 ∫
+
+−−
2
ba
2
ba3
2222
22
22)xb()ax(
xd.x
Q.32 Comment upon the nature of roots of the quadratic equation x2 + 2x = k + ∫ +1
0
dt|kt| depending on the
value of k + R.
Q.33 ∫a2
0
x sin−1
−a
xa2
2
1dx Q.34 Prove that
( )dx
x
dx
xn
nn
1 11
0
1
0 + −= ∫∫
∞
/ (n > 1)
Q.35 Show that ∫∫−− =
x
0
4z4xx
0
zxz dzeedze·e222
Q.36 ∫π
0
( )π−
π
x2
xcos.sin.x2sinx2
2
dx
Q.37 (a) ∫1
0 32 xxx
xd.
x1
x1
+++−
, (b) dxx
1x1n
1xx
1x2
51
124
2
−++−
+∫
+
l
Q.38 Show that dx
x x20 2 1+ +
∞
∫cosθ
= 2 dx
x x20
1
2 1+ +∫cosθ
=
θθ
θ π
θ πθ
θ π π
sin( , )
sin( , )
if
if
∈
−∈
L
N
MMMM
0
22
Q.39 ∫π
+
2
02
2
dxxsin8
xsinx
Q.40 ∫π
+−4
02
2
xdxcos)x2sin1(
)x2cosx2(sinxQ.41 Prove that ∫
x
0
f t dt
u
( )
0
∫
du = ∫x
0
f (u).(x − u) du.
Q.42 ∫π
0
dx
x( cos )5 4 2+Q.43 Evaluate ∫
1
0
ln ( )x1x1 ++− dx Q.44 ∫16
11xtan 1 −− dx
Q.45 ∫
−∞→
+n1
n1
2
ndx|x|)xcos2007xsin2006(nLim . Q.46 Show that f
a
x
x
a
x
xdx a f
a
x
x
a
dx
x( ).
lnln . ( ).
0 0
∞ ∞
∫ ∫+ = +
Q.47 Evaluate the definite integral, ∫− +
++1
1666
6911668998332
dxx1
)xsin·x4xx2(
Q.48 Prove that
(a) α
β
∫ )x()x( −βα− dx = ( )
8
2 πα−β(b)
α
β
∫x
x
−βα− dx = ( )
2
πα−β
(c) α
β
∫)x()x(x
xd
−βα− =
πα β
where α , β > 0 (d) α
β
∫ )x()x(
xd.x
−βα− = ( )2
πβ+α where α < β
Q.49 If f(x) =
4 1 1
1 1 1
1 1
2
2 2 2
2 2 2
cos
(cos ) (cos ) (cos )
(cos ) (cos ) cos
x
x x x
x x x
− + −+ +
, find f x dx( )
−
∫π
π
2
2
Q.50 Evaluate : ∫−−
1
0
1xtann dx)x(cossin·e1l
.
EXERCISE–3Q.1 If the derivative of f(x) wrt x is
)x(f
xcos then show that f(x) is a periodic function .
Q.2 Find the range of the function, f(x) = sin
cos
x dt
t x t1 2 21
1
− +−∫ .
Q.3 A function f is defined in [−1 , 1] as f′(x) = 2 x sin x
1 − cos
x
1; x ≠ 0 ; f(0) = 0;
f (1/π) = 0. Discuss the continuity and derivability of f at x = 0.
Q.4 Let f(x) = [ − − ≤ ≤− < ≤
1 2 0
1 0 2
if x
x if x and g(x) = ∫
−
x
2
f(t) dt. Define g (x) as a function of x and test the
continuity and differentiability of g(x) in (−2, 2).
Q.5 Prove the inequalities:
(a) π6
< dx
x x4
2
82 30
1
− −<∫
π(b) 2 e−1/4 < ∫
−2
0
xx2
e dx < 2e².
dx2
0
<+∫
π
then find a & b. (d) 2
1 ≤ ∫ +
2
02x2
dx ≤
6
5
Q.6 Determine a positive integer n ≤ 5, such that ∫1
0
ex (x − 1)n dx = 16 − 6e.
Q.7 Using calculus
(a) If x < 1 then find the sum of the series ∞++
++
++
++
......x1
x8
x1
x4
x1
x2
x1
18
7
4
3
2.
(b) If x < 1 prove that 284
73
42
3
2 xx1
x21......
xx1
x8x4
xx1
x4x2
xx1
x21
+++
=∞++−
−+
+−−
++−
−.
(c) a < b10 3cos x
(c) Prove the identity f (x)= tanx + 1
2tan
x
2 +
1
22tan
x
22 + .... + 1
2 1n− tanxn2 1− =
1
2 1n− cot xn2 1− – 2cot 2x
Q.8 If φ(x) = cos x − ∫x
0
(x − t) φ(t) dt. Then find the value of φ′′ (x) + φ(x).
Q.9 If y = dt)tx(asin·)t(fa
1x
0
−∫ then prove that yadx
yd 2
2
2
+ = f (x).
Q.10 If y = ∫x
1
tdtnl
x , find xd
yd at x = e.
Q.11 If f(x) = x + ∫1
0
Q.12 A curve C1 is defined by:
dx
dy = ex cos x for x ∈ [0, 2π] and passes through the origin. Prove that the
roots of the function (other than zero) occurs in the ranges 2
π < x < π and
2
3π < x < 2π.
Q.13(a) Let g(x) = xc . e2x & let f(x) = 0
x
∫ e2t . (3 t2 + 1)1/2 dt . For a certain value of 'c', the limit of ′′
f x
g x
( )
( )
as x → ∞ is finite and non zero. Determine the value of 'c' and the limit.
(b) Find the constants 'a' (a > 0) and 'b' such that, 0x
Lim→
t d ta t
bx x
x 2
0+
−
∫
sin = 1.
Q.14 Evaluate: ∫ +−+
+∞→
x3
x
1sin2
2
4
xdt
)3t)(3t(
1t3
dx
dLim
Q.15 Given that Un = {x(1 − x)}n & n ≥ 2 prove that
2n
2
xd
Ud = n (n − 1) U
n−2 − 2 n(2n − 1)U
n−1,
further if Vn = ∫
1
0
ex . Un dx, prove that when n ≥ 2, V
n + 2n (2n − 1).V
n−1− n (n − 1) V
n−2 = 0
Q.16 If ∫∞
0
dttx
tn22 +
�=
π �n 2
4 (x > 0) then show that there can be two integral values of ‘x’ satisfying this
equation.
Q.17 Let f(x) =
1 0 1
0 1 2
2 2 32
− ≤ ≤< ≤
− < ≤
x if x
if x
x if x( )
. Define the function F(x) = ∫x
0
f(t) dt and show that F is
continuous in [0, 3] and differentiable in (0, 3).
Q.18 Let f be an injective function such that f(x) f(y) + 2 = f(x) + f(y) + f(xy) for all non negative real
′ (0) = 0 & f ′ (1) = 2 ≠ f(0) . Find f(x) & show that, 3 ∫ f(x) dx − x (f(x) + 2) is a constant.
Q.19 Evaluate: (a) ∞→nLim
n/1
2
2
2
2
2
2
2 n
n1.....
n
31
n
21
n
11
+
+
+
+ ;
(b) ∞→nLim 1 1
1
2
2
3
4n n n
n
n++
++ +
..... ; (c) ∞→n
Limn/1
nn
!n
;
(d) Given
n1
nn2
nn3
n C
CLim
∞→=
b
a where a and b are relatively prime, find the value of (a + b).
[xy² + x²y] f(y) dy where x and y are independent variable. Find f(x).
x &
y with f
Q.20 Prove that sin x + sin 3x + sin 5x + .... + sin (2k − 1) x = xsin
xksin 2
, k ∈ N and hence
prove that , ∫π 2/
0 xsin
xksin2
dx = 1k2
1......
7
1
5
1
3
11
−+++++ .
Q.21 If Un= ∫
π 2/
0 xsin
xnsin2
2
dx , then show that U1 , U
2 , U
3 , ..... , U
n constitute an AP .
Hence or otherwise find the value of Un.
Q.22 Solve the equation for y as a function of x, satisfying
x · ∫∫ +=x
0
x
0
dt)t(y·t)1x(dt)t(y , where x > 0, given y (1) = 1.
Q.23 Prove that : (a) Im , n
= ∫1
0
xm . (1 − x)n dx = !)1nm(
!n!m
++ m , n ∈ N.
(b) Im , n
= ∫1
0
xm . (ln x)n dx = (−1)n 1n)1m(
!n++
m , n ∈ N.
Q.24 Find a positive real valued continuously differentiable functions f on the real line such that for all x
f 2(x) = ( ) ( )( )∫ +
x
0
22dt)t(')t( ff + e2
Q.25 Let f(x) be a continuously differentiable function then prove that, ∫x
1
[t] f ′ (t) dt = [x]. f(x) −∑=
]x[
1k
)k(f
where [. ] denotes the greatest integer function and x > 1.
Q.26 Let f be a function such that f(u) − f(v) ≤ u − v for all real u & v in an interval [a, b] . Then:(i) Prove that f is continuous at each point of [a, b] .
(ii) Assume that f is integrable on [a, b]. Prove that, f x dx b a f ca
b
( ) ( ) ( )− −∫ ≤ ( )b a− 2
2, where a ≤ c ≤ b
Q.27 Let F (x) = ∫−
+x
1
2 dtt4 and G (x) = ∫ +1
x
2 dtt4 then compute the value of (FG)' (0) where dash
denotes the derivative.
Q.28 Show that for a continuously thrice differentiable function f(x)
f(x) − f(0) = xf′ (0) + ′′f x( ).0
2
2
+ 1
2
2
0
′′′ −∫ f t x t dtx
( )( )
Q.29 Prove that k
n
=∑
0
(− 1)k ( )kn
1
1k m+ + =
k
m
=∑
0
(− 1)k ( )km
1
1k n+ +
(a) Prove that f (x) + g (x) = 6 for all x. (b) Find f (x) and g (x).
EXERCISE–4
Q.1 Find Limitn → ∞ S
n , if : S
n =
1
2
1
42
1
1
42
4
1
32
2 1n
n n n n
+−
+−
+ ++ −
.......... . [REE ’97, 6]
Q.30 Let f and g be function that are differentiable for all real numbers x and that have the followingproperties:(i) f ' (x) = f (x) – g (x) (ii) g ' (x) = g (x) – f (x)(iii) f (0) = 5 (iv) g (0) = 1
Q.2 (a) If g (x) = cos4
0
t dtx
∫ , then g (x + π) equals :
(A) g (x) + g (π) (B) g (x) − g (π) (C) g (x) g (π) (D) [ g (x)/g (π) ]
(b) Limitn → ∞
1
2 21
2
n
r
n rr
n
+=∑ equals :
(A) 1 5+ (B) − +1 5 (C) − +1 2 (D) 1 2+
(c) The value of π πsin ( ln )x
x
e
1
37
∫ dx is _______ .
(d) Let d
dxF x
e
x
x
( )sin
= , x > 0 . If 2
2
1
4e
x
xsin
∫ dx = F (k) − F (1) then one of the possible
values of k is ______.
(e) Determine the value of 2 1
12
x x
xdx
( sin )
cos
+
+−∫π
π. [JEE ’97, 2 + 2 + 2 + 2 + 5]
Q.3 (a) If f t dt x t f t dtx
x
( ) ( )= + ∫∫1
0
, then the value of f (1) is
(A) 1/ 2 (B) 0 (C) 1 (D) − 1/ 2
(b) Prove that tan tan− −
− +
=∫ ∫
1
20
11
0
11
12
x xdx x dx . Hence or otherwise, evaluate the integral
( )tan− − +∫1 2
0
1
1 x x dx [JEE’98, 2 + 8]
Q.4 Evaluate 1
5 2 2 12 2 40
1
( )( )( )+ − + −∫x x e
dxx
[REE ’98, 6 ]
Q.5 (a) If for al real number y, [y] is the greatest integer less than or equal to y, then the value of the
integral π
π
/
/
2
3 2
∫ [2 sin x] dx is :
(A) − π (B) 0 (C) − π2
(D) π2
(b) dx
x14
3 4
+∫cos
/
/
π
π is equal to :
(A) 2 (B) − 2 (C) 1
2(D) −
1
2
(c) Integrate :
( )x x
x x
3
22
3 2
1 1
+ +
+ +∫
( ) dx
(d) Integrate: ∫π
−+0
xcosxcos
xcos
dxee
e[JEE '99, 2 + 2 + 7 + 3 (out of 200)]
Q.6 Evaluate the integral 3 2 1
0
6 cos
cos
/ x
x
−∫
π
dx. [ REE '99, 6]
Q.7 (a) The value of the integral e
e
−∫
1
2
loge x
x d x is :
(A) 3/2 (B) 5/2 (C) 3 (D) 5
(b) Let g (x) = 0
x
∫ f (t) d t , where f is such that 1
2 ≤ f (t) ≤ 1 for t ∈ (0, 1] and 0 ≤ f (t) ≤
1
2for t ∈ (1, 2]. Then g (2) satisfies the inequality :
(A) − 3
2 ≤ g (2) <
1
2(B) 0 ≤ g (2) < 2 (C)
3
2 < g (2) ≤
5
2(D) 2 < g (2) < 4
(c) If f (x) = { e x for x
otherwise
xcos. sin | | ≤ 2
2. Then
−∫2
3
f (x)d x :
(A) 0 (B) 1 (C) 2 (D) 3
(d) For x > 0, let f (x) = 1
x
∫�n t
t1 + dt. Find the function f (x) + f (1/x) and show that,
f (e) + f (1/e) = 1/2 . [JEE 2000, 1 + 1 + 1 + 5]
Q.8 (a) Sn =
1
1 + n +
1
2 2+ n + ........ +
1
2n n+ . Find Limit
n → ∞ Sn .
(b) Given 0
1
∫sin t
t1 + d t = α , find the value of
4 2
4
π
π
−∫
sin t
t
2
4 2π + − d t in terms of α .
[ REE 2000, Mains, 3 + 3 out of 100]
Q.9 Evaluate sin− +
+ +
∫
1
2
2 2
4 8 13
x
x xdx .
Q.10 (a) Evaluate cos
cos sin
/ 9
3 30
2x
x xdx
+∫π
. (b) Evaluate xdx
x1+∫cos sinα
π
0
[ REE 2001, 3 + 5]
Q.11 (a) Let f(x) = 2 2
1
−∫ t dt
x
. Then the real roots of the equation x2 – f ′ (x) = 0 are
(A) +1 (B) + 1
2(C) +
1
2(D) 0 and 1
(b) Let T > 0 be a fixed real number. Suppose f is a continuous function such that for all x ∈ R
f (x + T) = f (x). If I = f( )x
T
0
∫ dx then the value of f(2 )x
T
3
3 3+
∫ dx is
(A) 3
2 I (B) 2 I (C) 3 I (D) 6 I
(c) The integral [ ]x nx
x+
+−
−
∫ l1
11
2
1
2
dx equals
(A) – 1
2(B) 0 (C) 1 (D) 2ln
1
2
[JEE 2002(Scr.), 3+3+3](d) For any natural number m, evaluate
x x x x x dxm m m m m m3 2 2
1
2 3 6+ + + +z c h c h , where x > 0 [JEE 2002 (Mains),4]
Q.12 If f is an even function then prove that ∫π 2
0
f (cos2x) cosx dx = ∫π 4
0
2 f (sin2x) cosx dx
[JEE 2003,(Mains) 2 out of 60]
Q.13 (a) ∫ +−1
0
dxx1
x1 =
(A) 12
+π
(B) 12
−π
(C) π (D) 1
(b) If 5t
0
t5
2dx)x(fx
2
=∫ , t > 0, then
25
4f =
(A) 5
2(B)
2
5(C) –
5
2(D) 1
[JEE 2004, (Scr.)]
(c) If ( ) θθ+θ
= ∫π
d.sin1
cos.xcosxy
2
16/2
x
2 then find
dx
dy at x = π. [JEE 2004 (Mains), 2]
(d) Evaluate dx
3|x|cos2
x43/
3/
3
∫π
π−
π+−
+π. [JEE 2004 (Mains), 4]
Q.14 (a) If ( )∫1
xsin
2 dt)t(ft = (1 – sin x), then f
3
1 is [JEE 2005 (Scr.)]
(A) 1/3 (B) 31 (C) 3 (D) 3
(b) ( )∫−
++++++0
2
23 dx)1xcos()1x(3x3x3x is equal to [JEE 2005 (Scr.)]
(A) – 4 (B) 0 (C) 4 (D) 6
(c) Evaluate: ∫π
+
0
|xcos| dxxsinxcos2
1cos3xcos
2
1sin2e . [JEE 2005, Mains,2]
Q.15 dx1x2x2x
1x
243
2
∫+−
− is equal to
(A) 2
24
x
1x2x2 +− + C (B)
3
24
x
1x2x2 +− + C
(C) x
1x2x2 24 +− + C (D)
2
24
x2
1x2x2 +− + C [JEE 2006, 3]
Comprehension
Q.16 Suppose we define the definite integral using the following formula ( ) ( ))b(f)a(f2
abdxxf
b
a
+−
=∫ , for
more accurate result for c ∈ (a, b) F(c) = ( ) ( ))c(f)b(f2
cb)c(f)a(f
2
ac+
−++
−. When c =
2
ba +,
( ) ))c(f2)b(f)a(f(4
abdxxf
b
a
++−
=∫
(a) ∫π 2/
0
dxxsin is equal to
(A) ( )218
+π
(B) ( )214
+π
(C) 28
π(D)
24
π
(b) If f (x) is a polynomial and if
( )
( )0
at
)a(f)t(f2
atdx)x(f
Lim3
t
a
at=
−
+−
−∫
→ for all a then the degree of f (x) can
atmost be(A) 1 (B) 2 (C) 3 (D) 4
(c) If f ''(x) < 0, ∀ x ∈ (a, b) and c is a point such that a < c < b, and (c, f (c) ) is the point lying on the curvefor which F(c) is maximum, then f '(c) is equal to
(A) ( ) ( )
ab
afbf
−−
(B) ( ) ( )( )
ab
afbf2
−−
(C) ( ) ( )
ab2
afbf2
−−
(D) 0
[JEE 2006, 5 marks each]
Q.17 Find the value of
( )
( )∫
∫
−
−
1
0
10150
1
0
10050
dxx1
dxx15050
[JEE 2006, 6]
ANSWER
EXERCISE–1
Q.1 ln
θθ++
2cos
2cos311 2
+ C Q.2 − x
x x
++ +
1
15 + c
Q.3 1
4ln(cos x + sin x) +
x
2 +
8
1 (sin 2x + cos 2x) + c Q.4
3
8 tan−1 x − ( )
x
x4 14 −−
3
16 ln
x
x
−+
1
1 + c
Q.5 2 tan−1 x x x+ + −
2 2 1 + c Q.6 x
e
e
xc
x x −
+
Q.7 (c) 2
1 (sin 2 θ) ln cos sin
cos sin
θ θθ θ
+−
−
2
1 ln (sec 2 θ) + c Q.8
2
1 ln
2xtan
xtan
+ + c
Q.9
+
+−
2
21
22 b
xtanatanx
ba
1+ c Q.10 2ln
1t2
t
+ +
1t2
1
+ + C when t = x + xx2 +
Q.11 3
1
2/32
2xx
++ − 2/1
2 2xx
2
++
+ c
Q.12 cos a . arc cos
acos
xcos − sin a . ln
−+ asinxsinxsin 22
+ c Q.13 ( )
3/8
2
)x(tan8
xtan413 +− + c
Q.14 2
1 ln
2
xtan +
4
1 sec²
2
x + tan
2
x + c Q.15 cxcosarcx12x1x ++−−−
Q.16 (a + x) arc tan a
x − xa + c Q.17
( )
+−++
23
22
x
11ln32.
x9
1x1x
Q.18 xln (lnx) − xnl
x + c Q.19 xx
x
ex1
1
ex1
exln
++
+ + c
Q.20 − ln (1 − x4) + c Q.21
−+++− − ttan)t1(n
2
1t
2
t
4
t6 12
24
l + C where t = x1/6
Q.22 4
2cos x
+ 2 tan−1 cos x2 − ln
1
1
2
2
+
−
cos
cos
x
x+ c Q.23 C – ln(1 + (x + 1)e–x) – xe)1x(1
1−++
Q24. sin−1
2
xsec
2
1 2 + c
Q.25 c)xcos3xsin34(
)xcos3xsin34(nl
24
1+
−−++
Q.26 2
1
π+−−
82
xtannl
2
1xcosxsin + c
Q.27 c)xcosx(sintanarcxcosxsin3
xcosxsin3nl
32
1+++
+−−+ Q.28 − − +
� � �n x n x n x(sec ) (sec ) (sec )
1
22
1
33 + c
Q.29 − 1
sinα ln [ ]cot cot cot cot cotx x x+ + + −α α2 2 1 + c Q.30 ln
xsinxcosx
xcosxsinx
−+
Q.31 c12
xtantanarc3x2 +
+− Q.32 cos
sin
2x
x − x − cot x . ln ( )( )e x xcos cos+ 2 + c
Q.33 ln (1 + t) − 1
4 ln (1 + t4) +
1
2 2 ln
t t
t t
2
2
2 1
2 1
− ++ +
− 1
2 tan−1 t2 + c where t = cotx
Q.34 c2
xtan
4
1
2
xtannl
2
1 2 +− Q.35 22 )1x(
xc
−−
Q.36 c – ecos x (x + cosec x) Q.37 sin− +
+1
2ax b
cxk Q.38 ex
x1
x1
−+
+ c Q.39 cx10x79
)20x7(2
2+
−−
−
Q.40 c1x
xnlxsecarc
2+
−− Q.41 �n
u
u u
uc where u
x
x
| |tan
2
4 2
12
31
13
1 2
3
1
1
−
+ ++
++ =
−+
−
Q.42 ( )8
3
1
2 5
5 1
5 11
1 1 2tan sin
− −+−+
− − −t nt
tx x� + c where t =
1
1
+−
x
x
Q.43 tan−1 2 2sin
sin cos
x
x x+
+ c Q.44 4 ln x +
x
7 + 6 tan–1(x) + 2x1
x6
+ + C
Q.45 c)1x(3
xtanarc
3
2+
+Q.46 −
− −+
− + − −
−+
−2 2
4
4 2 2 2 2 1
3
2 21x x
xn
x x x
x
x� sin + c
Q.47 cx
x.
2+
α−β−
β−α−
Q.48 2
1 ln
−
+++++ 122x
1x2
x
1x
2
+ C
Q.49
+−
−
−+
t1
t1n
2
1
t2
t2n
2
1ll where t = cosθ and θ = cosec–1(cotx)
Q.50
α
−
α −
2eccos
x2
1xtan·
2eccos
2
1 21
EXERCISE–2
Q.1 6
2π Q.2 2nl Q.3 6 − 2e Q.4
π2
1− Q.5 5
64
πQ.6
8
π ln 2
Q.7 1 – sec(1) Q.8 62 Q.9 2 2 + 4
3 ( )3 3 2 2− Q.12
22
7−
π Q.13
8
π(1 − ln 4)
Q.14 )12(n424 +− l Q.15 3
3π Q.16
22
)ba( +π Q.17
2
3
πQ.18 – )1e(
5
23 2 +π
Q.19 5
216
22−
πQ.20 0 Q.21
−
3
1tanarc
3
2tanarc
3
1
Q.22 ( )a bπ π+ 2
3 3Q.23
2
)3( +ππQ.24 5250 Q.27
36
2π
Q.28 )ba(a2 +
πQ.29
5
3
πQ.30
16
3 2πQ.31
12
πQ.32 real & distinct ∀ k ∈ R
Q.33 4
a2πQ.36
π8
Q.37 (a) 3
π; (b) 2n
8l
π Q.39 –
3
2 2π ln 2 Q.40
π2
16 −
π4
ln2
Q.42 27
5π Q.43
1
22
21ln + −
πQ.44 32
3
16−
πQ.45 2007 Q.47
666
4+π
Q.49 –2π – 32
15Q.50
2
1)2n1(
48
2
++π
−π
l
EXERCISE–3
Q.2 −
π π2 2
, Q.3 cont. & der. at x = 0
Q.4 g(x) is cont. in (−2 , 2); g(x) is der. at x = 1 & not der. at x = 0 . Note that ;