indefinite integration - arride learning€¦ · topic p n. teory 01 - 03 ecis - 1 04 - 11 ecis - 2...

Topic Page No.

Theory 01 - 03

Exercise - 1 04 - 11



Exercise - 4 18

Answer Key 19 - 23

Contents

INDEFINITE INTEGRATION

SyllabusIntegration as the inverse process of differentiation, indefinite integrals of

standard function, integration by parts, integration by the methods of

substitution and partial fractions

Name : ____________________________ Contact No. __________________

ARRIDE LEARNING ONLINE E-LEARNING ACADEMYA-479 indra Vihar, Kota Rajasthan 324005

Contact No. 8033545007

Page No. # 1Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005

KEY CONCEPTS

1. DEFINITION :If f & g are function of x such that g’ (x) = f (x) then the function g is called a PRIMITIVE ORANTIDERIVATIVE OR INTEGRAL of f(x) w.r.t. x and is written symbolically as

)x(f}c)x(g{dxdc)x(gdx)x(f =+Û+=ò , where c is called the constant of integration.

2. STANDARD RESULTS:

(i) 1nc)1n(a

)bax(dx)bax(1n

n -¹++

+=++

ò (ii) cbaxna1

baxdx ++=

+ò l

(iii) cea1dxe baxbax += ++ò (iv) c)0a(

naa

p1dxa

qpxqpx +>=

++ò l

(v) ò ++-=+ c)baxcos(a1dx)baxsin( (vi) ò ++=+ c)baxsin(

a1dx)baxcos(

(vii) ò ++=+ c)baxsec(na1dx)baxtan( l (viii) ò ++=+ c)baxsin(n

a1dx)baxcot( l

(ix) ò ++=+ c)baxtan(a1dx)bax(sec2 (x) ò ++-=+ c)baxcot(

a1dx)bax(eccos 2

(xi) c)bax(eccosa1dx)baxcot(.)bax(eccos ++-=++ò

(xii) c)xtanx(secndxxsec ++=ò l OR ò +÷øöç

èæ +p= c

2x

4tanndxxsec l

(xiii) ò +-= c)xcotxec(cosndxxeccos l OR ò += c2xtanndxxeccos l OR )xcotecx(cosn +- l

(xiv) ò +=-

- caxsin

xadx 1

22 (xv) ò +=+

- caxtan

a1

xadx 1

22

(xvi) ò +=-

- caxsec

a1

axxdx 1

22 (xvii) ò úûù

êëé ++=

+

2222

axxnax

dx l

(xviii) ò úûù

êëé -+=

-

2222

axxnax

dx l (xix) ò +-+=

-c

xaxan

a21

xadx

22 l

(xx) ò ++-=

-c

axaxn

a21

axdx

22 l (xxi) ò ++-=- - caxsin

2axa

2xdxxa 1

22222

INDEFINITE INTEGRATION


(xxii) ò +++=+ - caxsinh

2aax

2xdxax 1

22222 (xxiii) ò +--=- - c

axcosh

2aax

2xdxax 1

22222

(xxiv) ò +-+

= c)bxsinbbxcosa(ba

edxbxsin.e 22

axax

(xxv) ò +++

= c)bxsinbbxcosa(ba

edxbxcos.e 22

axax

3. TECHNIQUES OF INTEGRATION :(i) Substitution or change of independent variable.

Integral I = ò )x(f dx is changed to ò f )t('f))t((f dt, by a suitable substitution x = f (t) provided the

later integral is easier to integrate.

(ii) Integration by part : ò ò ò ò úûù

êëé-= dxv.dxdudxvudxv.u dx where u & v are dif ferentiable func-

tion . Note: While using integration by parts, choose u & v such that

(a) ò dxv is simple & (b) ò ò úûù

êëé dxv.dxdu

dx is simple to integrate.

This is general ly 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, splitting a bigger fraction into smaller fraction by known methods.

4. INTEGRALS OF THE TYPE :

(i) [ ]ò dx)x('f)x(f nOR ò n)]x(f[

)x('f dx put f(x) = t & proceed.

(ii) òòò ++++++

dxcbxax,cbxax

dx,cbxax

dx 222

Express ax2 + bx + c in the from of perfect square & then apply the standard results.

(iii) òò ++

+++

+ dxcbxax

qpx,dxcbxax

qpx22

Express px + q = A (differential coefficient of denominator ) + B.

(vi) ò +=+ c)x(f.edx)]x('f)x(f[e xx (v) ò +=+ c)x(fxdx)]x('xf)x(f[

(vi) ò Î+

Nn)1x(x

dxn Take xn common & put 1 + x–n = t.

(vii)( )ò Î

+- Nn

1xx

dx

n)1n(

n2, take xn common & put 1 + x–n = tn


(viii)( )ò +

n/1nn x1x

dx take xn common as x and put 1 + x–n = t.

(ix) ò + xsinbadx

2 OR ò + xcosbadx

2 OR ò ++ xcoscxcosxsinbxsinadx

22

Multiply N’ & D’ by sec2 x & put tan x = t .

(x) ò + xsinbadx

OR ò + xcosbsadx

OR ò ++ xcoscxsinbadx

Hint : Conv er t sines & cosines into thei r respect i v e tangents of ha l f the angles,

put tan t2x

=

(xi) ò ++++ .dx

nxsin.mxcos.cxsin.bxcos.a

l Express Nr º A (Dr) + B dxd

(Dr) + c & proceed.

(xii) ò +++ dx

1Kxx1x24

2OR ò ++

- dx1Kxx

1x24

2where K is any constant.

Hint: Divide Nr & Dr by x2 & proceed.

(xiii) ò ++ qpx)bax(dx

& ò =++++

22 tqpxput;

qpx)cbxax(dx

(xiv) ò =++++

;t1baxput,

rqxpx)bax(dx

2 ò ++ rpx)cax(dx

22 , put x = t1

(xv) ò -ba-x

x dx or ò -ba- )x)(x( ; put x = a cos2 q + b sin2 q

ò b-a-

xx

dx or ò b-a- )x)(x( ; put x = a sec2 q – b tan2 q

ò b-a- )x)(x(dx

; put x – a = t2 or x – b = t2.


PART - I : OBJECTIVE QUESTIONS

* Marked Questions are having more than one correct option.

Section (A) : Integration Using Standard Integral

A-1. The value of ò a+ )xsin(.xsindx

is equal to

(A) cosec a ln C)xsin(

xsin+

a+ (B) cosec a ln Cxsin

)xsin(+

a+

(C) cosec a ln Cxsec

)xsec(+

a+(D) cosec a ln C

)xsec(xsec

+a+

A-2.1 xdx tan a b

1 sinx 2æ ö= + +ç ÷+ è øò , then

(A) a ,b R4p

= - Î (B) a ,b R4p

= Î (C) 5a ,b R4p

= Î (D) none of these

A-3. If 1(sin2x cos2x) dx sin(2x a) b,2

- = - +ò then

(A) 5a ,b R4p

= Î (B) 5a ,b R4p

= - Î (C) a ,b R4p

= Î (D) none of these

A-4. The value of cos2x dxcosxò is equal to

(A) 2 sin x – l n |sec x + tan x| + C (B) 2 sin x – l n |sec x – tan x| + C(C) 2 sin x + l n |sec x + tan x| + C (D) None of these

Section (B) : Integration Using Substitution

B-1. The value of xa dxxò is equal to

(A) xa Cx

+ (B) x2a C

na+

l(C) x2a . n a C+l (D) none of these

B-2. The value of x55 5.5.5xx5

ò dx is equal to

(A) C)5n(

53

5x

+l

(B) C)5n(5 35x5

+l (C) C)5n(

53

5x5

+l

(D) none of these

B-3. The value of ò xcosxsinxtan

dx is equal to

(A) Cxtan2 + (B) Cxcot2 + (C) C2

xtan+ (D) none of these


B-4. If ò- x

x

41

2dx = K sin–1 (2x) + C, then the value of K is equal to

(A) ln 2 (B) 2n21l (C)

21

(D) 2n1l

B-5. If y = ò + 2/32 )x1(dx

and y = 0 when x = 0, then value of y when x = 1, is

(A) 32

(B) 2 (C) 23 (D) 21

B-6. The value of ò tan3 2 x sec 2 x dx is equal to :

(A) 13 sec3 2 x –

12

sec 2 x + C (B) – 16

sec3 2 x – 12

sec 2 x + C

(C) 16 sec3 2 x –

12

sec 2 x + C (D) 13 sec3 2 x + 1

2sec 2 x + C

B-7. The value of ò + 2)xcosx(sinx2cos

dx is equal to :

(A) xcosxsin1

+-

+ C (B) ln (sin x + cos x) + C

(C) ln (sin x – cos x) + C (D) ln (sin x + cos x)2 + C

B-8. The value of ò a++ )]xtan(.xtan1[ dx is equal to :

(A) cos a . ln )xsin(xsin

a+ + C (B) tan a . ln )xsin(xsin

a+ + C

(C) cot a . ln xsec)xsec( a+

+ C (D) cot a . ln xcos)xcos( a+

+ C

B-9*. The value of mx nx2 . e dxò (when m, n Î N) is equal to :

(A) mx nx2 3 C

m n2 n n3+

++l l

(B) (m n2 n n3)xe

n n2 n n3

+

+

l l

l l+C (C) ( )

mx nx

m n

2 .3 Cn 2 ,3

+l

(D) x x(mn).2 .3 C

m n2 n n3+

+l l

Section (C) : Integration by Parts

C-1. The value of ò -- xe)1x( dx is equal to :

(A) –xex + C (B) xex + C (C) – xe–x + C (D) xe–x + C

C-2. The value of ò- xtan 1

e ÷÷ø

öççè

æ

+++

2

2

x1xx1

dx is equal to :

(A) x xtan 1e

- + C (B) x2 xtan 1

e-

+ C (C) x1

xtan 1e

- + C (D) none of these


C-3. The value of ò ¢¢-¢¢ )]x(g)x(f)x(g)x(f[ dx is equal to :

(A) )x(g)x(f

¢ (B) f¢(x) g(x) – f(x) g¢(x)

(C) f(x) g¢(x) – f¢(x) g(x) (D) f(x) g¢(x) + f¢(x) g¢(x)

C-4*. If 3x 3xe cos 4xdx e (A sin4x Bcos4x) C= + +ò then :

(A) 4A = 3B (B) 2A = 3B (C) 3A = 4B (D) 4A + 3B = 1

Section (D) : Algebraic Integral

D-1. The value of ò ++ 1xxdx

2 is equal to :

(A) 23

tan–1 ÷÷ø

öççè

æ +

31x2

+ C (B) 32

tan–1 ÷÷ø

öççè

æ +

31x2

+ C

(C) 31

tan–1 ÷÷ø

öççè

æ +

31x2

+ C (D) none of these

D-2. The value of ò + 4/342 )1x(x1

dx is equal to

(A) Cx11

4/1

4 +÷ø

öçè

æ + (B) (x4 + 1)1/4 +C (C) Cx11

4/1

4 +÷ø

öçè

æ - (D) Cx11

4/1

4 +÷ø

öçè

æ +-

D-3. The value of 3

dxx 1 x-

ò is equal to

(A) 3

3

1 1 x 1n C3 1 x 1

- -+

- +l (B)

21 1 x 1n C3 1 x2 1

- ++

- -l (C) 3

1 1n C3 1 x

+-

l (D) 31 n 1 x C3

- +l

D-4. The value of ò +

-

1e1e

x

x

dx is equal to :

(A) ln ÷øöç

èæ -+ 1ee x2x – sec–1 (ex) + C (B) ln ÷

øöç

èæ -+ 1ee x2x + sec–1 (ex) + C

(C) ln ÷øöç

èæ -- 1ee x2x – sec–1 (ex) + C (D) none of these

D-5. If ò + 34 xxdx

= 2xA

+ xB

+ ln 1xx+ + C, then :

(A) A = 21

, B = 1 (B) A = 1, B = – 21

(C) A = – 21

, B = 1 (D) none of these


Section (E) : Integration of Trigonometric Functions:

E-1. If ò xsinxcos

11

3

dx = – 2÷÷

ø

ö

çç

è

æ+

--

xtanBxtanA 25

29

+ C, then :

(A) A = 91

, B = 51-

(B) A = 91

, B = 51

(C) A = – 91

, B = 51

(D) none of these

E-2. The value of ò -1xsec dx is equal to :

(A) 2 ln ÷÷ø

öççè

æ-+

21

2xcos

2xcos 2

+ C (B) ln ÷÷ø

öççè

æ-+

21

2xcos

2xcos 2

+ C

(C) – 2 ln ÷÷ø

öççè

æ-+

21

2xcos

2xcos 2

+ C (D) none of these

E-3. The value of 3

dxcos x sin2xò is equal to

(A) 5 / 212 cosx tan x C

5æ ö+ +ç ÷è ø

(B) 5 / 212 tan x tan x C

5æ ö+ +ç ÷è ø

(C) 5 / 212 tan x tan x C

5æ ö- +ç ÷è ø

(D) none of these

Section (F) : Miscellaneous

F-1. If x x

x x

4e 6e dx9e 4e

-

-

+-ò = Ax + B nl |9e2x – 4| + C, then

(A) 3 35A ,F ,C 02 36

= - = = (B) 35 3A , B ,C R36 2

= = - Î

(C) 3 35A , B ,C R2 36

= - = Î (D) 3 35A , B ,C R2 36

= = Î

F-2*. Let f' (x) = 3x2 sin 1x

– x cos 1x

, if x ¹ 0 ; f(0) and f(1/p) = 0 then ;

(A) f(x) is continuous at x = 0 (B) f(x) is non derivable at x = 0(C) f' (x) is continuous at x = 0 (D) f' (x) is non derivable at x = 0

PART - II : SUBJECTIVE QUESTIONS

Section (A) : Integration Using Standard Integral

A-1. Integrate with respect to x:(i) (2x + 3)5 (ii) sin 2x (iii) sec2 (4x + 5) (iv) sec (3x + 2) (v) tan (2x + 1)

(vi) 23x + 4 (vii) 1

2x 1+(viii) e4x + 5 (ix) x 1+ (x)

12x 1+


A-2. Integrate with respect to x :

(i) sin 2x + 1

x 1+(ii) tan (3x + 1) + e4x + 5 (iii) 2 tan (4x + 5) (iv) sin2 x

(v) cos2x (vi) sin 2x cos 3x (vii) 1

x 3 x 2+ - +

Section (B) : Integration Using Substitution

B-1. Integrate with respect to x :

(i) x sin x2 (ii) 2

xx 1+

(iii) sec2 x tan x (iv) x

x

e 1e x

++

(v) 1 sinxx cosx

-+

(vi) 2x

2x

ee 2-

(vii) 2

cos2x x 1x sin2x 2x

+ ++ +

(viii) sec x

n(sec x tanx)+l(ix)

xx 2+

(x) 2

xx

1ee

æ ö+ç ÷è ø(xi) (ex + 1)2 ex (xii) ( )5

1x x 1+

Section (C) : Integration by PartsC-1. Integrate with respect to x :

(i) x sin x (ii) x l n x (iii) x sin2x (iv) x tan–1 x (v) l nx (vi) sec3x

(vii) 23 x2x e (viii) 1sin x- (ix) 2 1

2

x tan x1 x

-

+(x) ex sin x (xi) ex (sec2x + tan x)

Section (D) : Algebraic IntegralD-1. Integrate with respect to x:

(i) 2x 4+ (ii) 2

1x 4+

(iii) 2

1x 4-

(iv) 2

1x 5+

(v) 2x 2x 5+ + (vi) 2

1x 2x 5+ +

(vii) 2(x 1) 1 x x- - - (viii) 2

2x 1x 3x 4

++ +

(ix) 5 3 3x a x+ (x) 15 5 5

1

x (1 x )+(xi)

2

4

x 8x

- (xii)3

3

x 1x x

-+

D-2. Integrate with respect to x :

(i) 1

(x 1)(x 2)+ + (ii) 2

1(x 1)(x 3)+ + (iii) 2

1(x 1) (x 2)+ + (iv)

1(x 1)(x 2)(x 3)+ + +

Section (E) : Integration of Trigonometric Functions:E-1. Integrate with respect to x :

(i) 1

2 cosx+(ii)

12 cosx-

(iii) 2sinx 2cosx3cos x 2sinx

++

(iv) 1

1 sinx cosx+ +

(v) 2

12 sin x+

(vi) 2cosec x.sinx

(sinx cosx)- (vii) 4

2

sin xcos x

Section (F) : MiscellaneousF-1. Integrate with respect to x :

(i) 4 2

1x x 1+ +

(ii) 2

4

1 x1 x

++

(iii) 2

2 4

1 x1 x x

-- +

F-2. Integrate with respect to x :

(i) 1

(x 1) x 2+ + (ii) 2

1(x 4) x 1- + (iii) 2

1(x 1) x 2+ + (iv) 2 2

1(x 1) x 2+ +


PART - III : MISCELLANEOUS OBJECTIVE QUESTIONS

Match The column-I

1. Column – I Column – II

(A) If F(x) = x sinx1 cos x

++ò dx and F(0) = 0, (p)

2p

then the value of F(p/2) is :

(B) Let F(x) = 1sin x

2

xe 11 x

- æ ö-ç ÷ç ÷-è ø

ò dx and F(0) = 1, (q) 3p

If F(1/2) = p

p 6/e3k , then the value of k is :

(C) Let F(x) = 2 2dx

(x 1) (x 9)+ +ò and F(0) = 0, (r)4p

if F( 3 ) = 365

k, then the value of k is :

(D) Let F(x) = tanx

sin xcosxò dx and F(0) = 0 (s) p

if F(p/4) = pk2

, then the value of k is :

2. If I = dx

a b cosx+ò , where a, b > 0 and a + b = u, a – b = v, then match the following column.

Column – I Column – II

(A) v = 0 (p) I = uv1

ln

2xtanv–u

2xtanvu +

+ C

(B) v > 0 (q) I = uv2

tan–1 ÷÷ø

öççè

æ

2xtan

uv

+ C

(C) v < 0 (r) I = vu–

1 ln

2xtanv––u

2xtanv–u +

+ C

(s)u2 tan

2x + C


Comprehension #1

Let In, m = n msin xcos x.dx.ò Then we can relate In, m with each of the following

(i) I n – 2 , m (ii) I n + 2 , m (iii) I n , m – 2 (iv) I n , m + 2

(v) I n–2 , m + 2 (vi) I n + 2 , m – 2

Suppose we want to establish a relation between In, m and In, m –2 , then we setP(x) = sinn + 1 x cosm – 1 x ..... (1)

In In, m and In , m – 2 the exponent of cos x is m and m – 2 respectively, the minimum of the two is m – 2, adding 1to the minimum we get m – 2 + 1 = m – 1 . Now choose the exponent m – 1 of cos x in P(x). Similarly choosethe exponent of sin x for P(x)Now differentiating both sides of (1), we getP' (x) = (n + 1) sinnx cosmx – (m – 1) sinn + 2 x cos m– 2 x

= (n + 1) sinnx cosmx – (m – 1) sinn x (1 – cos2x) cos m– 2 x= (n + 1) sinnx cosmx – (m – 1) sinn x cosm– 2 x (m –1) sinnx cosmx= (n + m) sinnx cosmx – (m – 1) sinn x cosm– 2 x

Now integrating both sides , we getsinn + 1 x cos m–1 x = (n + m) In, m – (m – 1) In , m – 2 .Similarly we can establish the other relations.

3. The relation between I4, 2 and I2, 2 is

(A) I4, 2 = 16 (– sin3x cos3x + 3I2, 2) (B) I4, 2 =

16 (sin3x cos3x + 3I2 , 2)

(C) I4 , 2 = 16 (sin3x cos3x – 3I2, 2) (D) I4 , 2 =

14

(– sin3x cos3x + 2I2 , 2)


(A) I 4, 2 = 15 (sin5x cos3x + 8I6, 2) (B) I4, 2 =

15 (– sin5x cos3x + 8I6 , 2)

(C) I4, 2 = 15 (sin5x cos3x – 8I6, 2) (D) I 4, 2 =

16 (sin5x cos3x + 8I6 , 2)


(A) I4, 2 = 13 (sin5 x cos3x + 8I4, 4) (B) I4, 2 =

13 (– sin5 x cos3x + 8I4, 4)

(C) I4, 2 = 13 (sin5 x cos3x – 8 I4, 4) (D) I4, 2 =

13 (sin5 x cos3x + 6 I4, 4)

Comprehension #2

It is known that

sinx cosx if 0 x2cos x sinxtanx cot x

sinx cosx 3if x2cosx sinx

ì p+ < <ï

ï+ = í- - pï + p < <ï - -î

( ) ( )d 1 3tanx cot x tanx cot x (tanx cot x), x 0, ,dx 2 2 2

p pæ ö æ ö- = + + " Î È pç ÷ ç ÷è ø è ø

and ( ) ( )d 1 3tanx cot x tanx cot x (tanx cot x), x 0, ,dx 2 2 2

p pæ ö æ ö+ = - + " Î È pç ÷ ç ÷è ø è ø


6. Value of integral I = ( ) 3tanx cot x dx where x 0, ,2 2p pæ ö æ ö+ Î È pç ÷ ç ÷

è ø è øò is

(A) 1 tanx cot x2 tan C

2- æ ö-

+ç ÷ç ÷è ø

(B) 1 tanx cot x2 tan C

2- æ ö+

+ç ÷ç ÷è ø

(C) 1 tanx cot x2 tan C

2- æ ö-

- +ç ÷ç ÷è ø

(D) 1 tanx cot x2 tan C

2- æ ö+

- +ç ÷ç ÷è ø

7. Value of the integral I = ( )tanx cot x dx,where x 0,2pæ ö+ Î ç ÷

è øò , is

(A) 12 sin (cosx sinx) C- - + (B) 12 sin (sinx cosx) C- - +

(C) 12 sin (sinx cosx) C- - + (D) 12 sin (sinx cosx) C-- + +

8. Value of the integral I = ( )tan x cot x dx+ò , where x Î 3,2pæ öpç ÷

è ø is

(A) 2 sin–1 (cos x – sin x) + C (B) 2 sin–1 (sin x – cos x) + C

(C) 2 sin–1 (sin x + cos x) + C (D) – 2 sin–1 (sin x + cos x) + C

ASSERTION/REASONING

9. STATEMENT-1 : 5(sin x)ò cos x dx = 6

xsin6

+ C.

STATEMENT-2 : n(f(x)) f (x)¢ò = 1n

))x(f( 1n

+

+

+ C, n Î I.

(A) Statement -1 is true, Statement - 2 is true ; Statement - 2 is correct explanation for Statement - 1(B) Statement -1 is true, Statement - 2 is true ; Statement - 2 is NOT correct explanation for Statement - 1(C) Statement -1 is true, Statement - 2 is false.(D) Statement -1 is false, Statement - 2 is true.

10. STATEMENT-1 : If x > 0, x ¹ 1 then 2x x(log e (log e) ) dx-ò = x logxe + C.

STATEMENT-2 : xe (f(x) f (x)) dx+ ¢ò = ex f(x) + C and et = x iff t = lnx.


11. STATEMENT-1 : x

xn (e 1)

e+

òl

dx = x – ÷÷ø

öççè

æ +x

x

ee1

ln (ex + 1) + C.

STATEMENT-2 : f (x)f(x)¢

ò dx = ln |f(x)| + C.



PART - I : OBJECTIVE QUESTIONS

1. If f(x) = ò-3x

x2sinxsin2dx, where x ¹ 0, then 0x

Limit® f'(x) has the value

(A) 0 (B) 1 (C) 2 (D) not defined

2. If ò -+

xtanxcot1x4cos

dx = A cos 4x + B; where A & B are constants, then

(A) A = – 1/4 & B may have any value (B) A = – 1/8 & B may have any value(C) A = – 1/2 & B = – 1/4 (D) none of these

3. The value of ( )xxx

e x+ò dx is equal to

(A) C]1xx[e2 x ++- (B) C]1x2x[e2 x ++-

(C) C]1xx[e2 x ++- (D) C]1xx[e2 x +++

4. The value of tane (sec sin )dq q - q qò is equal to

(A) tane sin Cq- q + (B) tane sin Cq q + (C) tane sec Cq q + (D) tane cos Cq q +

5. dx)x1(x

x17

7

ò +- equals :

(A) 72n| x | n|1 x | c7

+ + +l l (B) 72n| x | n|1 x | c4

- - +l l

(C) 72n| x | n |1 x | c7

- + +l l (D) 72n| x | n|1 x | c4

+ - +l l

6.1 cosx

cos cos x-a -ò dx where 0 < a < x < p , equals :

(A) c2xcos

2cosn2 +÷ø

öçèæ -

al (B) 1

xcos22 cos c

cos2

-

æ öç ÷

+ç ÷aç ÷ç ÷è ø

(C) c2xcos

2cosn22 +÷ø

öçèæ -

al (D) 1

xcos22sin c

cos2

-

æ öç ÷

- +ç ÷aç ÷ç ÷è ø


7. The value of ò +- 4/153 ]2x()1x[(1

dx is equal to

(A) C2x1x

34 4/1

+÷ø

öçè

æ+-

(B) C1x2x

34 4/1

+÷ø

öçè

æ-+

(C) C2x1x

31 4/1

+÷ø

öçè

æ+-

(D) C1x1x

31 4/1

+÷ø

öçè

æ-+

8. The value of n sin x(x e cos x)-ò l dx is iqual to

(A) x cos x + C (B) sin x – x cos x + C(C) – e lnx cos x + c (D) sin x + x cos x + C

9. Antiderivative of 2

2

sin x1 sin x+

w.r.t. x is :

(A) ( )2x arctan 2 tanx C2

- + (B) 1 tanxx arctan C2 2

æ ö- +ç ÷

è ø

(C) ( )x 2 arctan 2 tanx C- + (D) tanxx 2 arctan C

2æ ö

- +ç ÷è ø

10. The value of dx2x3cos

2xcosxsin4ò is equal to

(A) Cx3cos31x2cos

21xcos ++- (B) Cx3cos

31x2cos

21xcos +--

(C) Cx3cos31x2cos

21xcos +++ (D) Cx3cos

31x2cos

21xcos +-+

11. The value of 1 x1 x

-

+ò dx is equal to

(A) ( )1x 1 x 2 1 x cos x C-- - - + + (B) ( )1x 1 x 2 1 x cos x C-- + - + +

(C) ( )1x 1 x 2 1 x cos x C-- - - - + (D) ( )1x 1 x 2 1 x cos x C-- + - - +

12. The value of ò )x16cos.x8cos.x4cos.x2cos.xcos.x(sin dx is equal to

(A) C1024

x16sin+ (B) C

1024x32cos

+- (C) C1096

x32cos+ (D) C

1096x32cos

+-

13. The value of 6 6

1cos x sin x+ò dx is equal to

(A) tan–1 (tan x + cos x) + C (B) – tan–1 (tan x + cos x) + C(C) tan–1 (tan x – cot x) + C (D) – tan–1 (tan x – cot x) + C

14. The value of xln(1 sinx) x tan

4 2ì üpæ ö+ + -í ýç ÷

è øî þò dx is equal to:

(A) x l n (1 + sin x) + C (B) l n (1 + sin x) + C(C) – x l n (1 + sin x) + C (D) l n (1 – sin x) + C


15. The value of ò +-

2x1.

1x1x

dx is equal to

(A) Cx

1xx1sin

21 +

-+- (B) C

x1cos

x1x 1

2++

- -

(C) Cx

1xxsec2

1 +-

-- (D) Cx

1x1xtan2

21 +-

-+-

16. If I = 3sinx sin x dx

cos2x+

ò = A cos x + B l n |f(x)| + C, then

(A) 1 1 2 cos x 1A , B , f(x)4 2 2 cos x 1

- -= = =

+(B)

1 3 2 cos x 1A , B , f(x)2 4 2 2 cos x 1

- -= - = =

+

(C) 1 3 2 cos x 1A , B , f(x)2 2 2 cos x 1

+= - = =

-(D)

1 3 2 cos x 1A , B , f(x)2 4 2 2 cos x 1

- -= = =

+

17. If 3

3 5

dx a cot x b tan x Csin x cos x

= + +ò , where C is an arbitary constant of integration, then the

values of 'a' and 'b' are respectively:

(A) –2 & 23 (B) 2 & –

23 (C) 2 &

23 (D) none

More than one choice tyep

18. If 1dx xItan mtan C

5 4cosx 2- æ ö= +ç ÷+ è øò then :

(A) I = 2/3 (B) m = 1/3 (C) I = 1/3 (D) m = 2/3

19. The value of 2 2

2

x cos x1 x++ò cosec2x dx is equal to;

(A) cot x – cot–1 x + C (B) C – cot x + cot–1 x

(C) – tan–1 x – cosec x Csec x

+ (D) 1n tan xe cot x C-

- - +l

20. The value of 4 4sin2x dx

sin x cos x+ò is equal to :

(A) ( ) Cxcotcot 21 +- (B) ( ) Cxtancot 21 +- -

(C) ( ) Cxtantan 21 +- (D) ( ) Cx2costan 1 +- -


21. The value of 2

dx

x x-ò is equals to

(A) cxsin2 1 +- (B) c)1x2(sin 1 +--

(C) )1x2(cos2c 1 -- - (D) cxx2cos 21 +--

22. The value of 2

x 1nx 1

x 1

-æ öç ÷+è ø

-òl

dx is equal to

(A) 21 x 1n C2 x 1

-+

+l (B) 21 x 1n C

4 x 1-

++

l (C) 21 x 1n C2 x 1

++

-l (D) 21 x 1n C

4 x 1+

+-

l

23. The value of n (tan x) dxsin xcosxòl is equal to

(A) 21 n (cot x) C2

+l (B) 21 n (sec x) C2

+l (C) 21 n (sinx sec x) C2

+l (D) 21 n (cos x cosec x) C2

+l

24. If In = ncot x dxò and I0 + I1 + 2 (I2 + .... + I8) + I9 + I10 = A 2 9u uu .......

2 9æ ö

+ + =ç ÷è ø

+ C, where u = cot x and

C is an arbitrary constant, then(A) A is constant (B) A = – 1 (C) A = 1 (D) A is dependent on X

PART - II : SUBJECTIVE QUESTIONS

Evaluate the following integrals:

1.1 dx

sin(x a)cos(x b)- -ò 2. tan x. tan2x. tan3x dxò 3. 3 3

x dxa x-ò

4.2 2

2 2

a xx dxa x

-+ò 5. ( )3 / 22

x nx dxx 1-

òl

6. 1 xsin dxa x

-

+ò

7. ò3

sin x2

xcos x sinxe .cos x

- dx 8. x x 1 dx

x 2+ +

+ò 9. 2

2sin2 cos6 cos 4sin

f - f- f - fò

10.2

6

4 x dxx+

ò 11. 4 4sin xcos x dxò 12. 4

1 dx1 sin x-ò

13.2

4 2

(x 1) dxx x 1

-+ +ò 14. ( )2 2sinx

1 xcosx dxx 1 x e

+

-ò


PART-I IIT-JEE (PREVIOUS YEARS PROBLEMS)

1. Integrate, ( )

3

22

x 3x 2 dxx 1 (x 1)

+ +

+ +ò [IIT-JEE 1999], Part-2]

2. Let f(x) = xe (x 1)(x 2)- -ò dx then f decreases in the interval : [IIT-JEE 2000, ]

(A) (– ¥, 2) (B) (–2, –1) (C) (1, 2) (D) ),2( ¥+

3. Evaluate : 12

2x 2sin dx4x 8x 13

- æ ö+ç ÷ç ÷+ +è ø

ò . [IIT-JEE 2001]

4. For any natural number m, evaluate ( ) ( )1m3m 2m m 2m mx x x 2x 3x 6+ + + +ò dx , where x > 0.

[IIT-JEE 2002]

5.2

3 4 2

x 1 dxx 2x 2x 1

-

- +ò is equal to : [IIT-JEE 2006]

(A) 2 2 14 2

2x x

xc- +

+ (B) 2 2 14 2

3x x

xc- +

+ (C) 2 2 14 2x xx

c- ++ (D)

2 2 12

4 2

2x x

xc- +

+

6. Let f x x

xn nb ge j

=+1

1/ for n ³ 2 and gx f f f xf occurs n times

b g b g b g= ° ° °..... .1 244 344

Then ( )n 2x g x dx-ò equals :

[IIT-JEE 2007]

(A) 1

11

1 1

n nnx Kn n

-+ +

-

b g e j (B) 1

11

1 1

nnx Kn n

-+ +

-

b g e j

(C) 1

11

1 1

n nnx Kn n

-+ +

+

b g e j (D) 1

11

1 1

nnx Kn n

-+ +

+

b g e j

7. Let F(x) be an indefinite integral of sin .2 x [IIT-JEE 2007]Statement - 1 : The function F(x) satisfies F x F x+ =pb g b g for all real x.because

Statement - 2 : sin sin2 2x x+ =pb g for all real x.(A) Statement - 1 is True, Statement - 2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement - 1 is True, Statement - 2 is True; Statement - 2 is NOT a correct explanation for Statement - 1.(C) Statement - 1 is True, Statement - 2 is False.(D) Statement - 1 is False, Statement - 2 is True.

8. Let x

4x 2xeI dx

e e 1=

+ +ò , x

4x 2xeJ dx

e e 1

-

- -=+ +ò . Then for an arbitrary constant C, the value of J – I

equals : [IIT-JEE 2008]

(A) 4x 2x

4x 2x1 e e 1log2 e e 1

æ ö- +ç ÷

+ +è ø + C (B)

2x x

2x x1 e e 1log2 e – e 1

æ ö+ +ç ÷

+è ø + C

(C) 2x x

2x x1 e e 1log2 e e 1

æ ö- +ç ÷

+ +è ø + C (D)

4x 2x

4x 2x1 e e 1log2 e e 1

æ ö+ +ç ÷

- +è ø + C


PART-II AIEEE (PREVIOUS YEARS PROBLEMS)

1. n

dxx(x 1)+ò is equal to : [AIEEE 2002]

(A) n1

log 1xxn

n

+ + c (B) n1

log n

n

x1x +

+ c (C) log 1xxn

n

+ + c (D) none of these

2. If ò a)–xsin(xsin

dx = Ax + B log sin (x – a) + c, then value of (A, B) is : [AIEEE 2004]

(A) (sin a, cos a) (B) (cos a, sin a) (C) (– sin a, cos a) (D) (– cos a, sin a)

3. ò xsin–xcosdx

is equal to : [AIEEE 2004]

(A) 21

log ÷øö

çèæ p

8–

2xtan + c (B)

21

log ÷øö

çèæ

2xcot + c

(C) 21

log ÷øö

çèæ p

83–

2xtan + c (D)

21

log ÷øö

çèæ p

+83

2xtan + c

4. ò ïþ

ïýü

ïî

ïíì

+

2

2)x(log1)1–x(log

dx is equal to : [AIEEE 2005]

(A) 1)x(log

x2 +

+ c (B) 2

x

x1xe+

+ c (C) 1x

x2 +

+ c (D) 1)x(logxlog2 +

5. ò + xsin3xcosdx

equals : [AIEEE 2007]

(A) 21

log tan ÷øö

çèæ p

+122

x + c (B)

21

log tan ÷øö

çèæ p

12–

2x

+ c

(C) log tan ÷øö

çèæ p

+122

x + c (D) log tan ÷

øö

çèæ p

12–

2x

+ c

6. The value of 2 ò÷øö

çèæ p

4–xsin

dxxsin is : [AIEEE 2008]

(A) x + log ÷ø

öçè

æ p4

–xcos + c (B) x – log ÷ø

öçè

æ p4

–xsin + x

(C) x + log ÷ø

öçè

æ p4

–xsin + c (D) x – log ÷ø

öçè

æ p4

–xcos + c

7. If the integral 5 tanx

sin x – 2cosxò dx = x + a ln | sinx – 2 cosx | + k, then a is equal to [AIEEE 2012]

(A) – 1 (B) – 2 (C) 1 (D) 2


NCERT BOARD QUESTIONS

Verify the following:

1.2x 1 dx2x 3

-+ò = x – log |(2x + 3)2| +C 2. 2

2x 3 dxx 3x

++ò = log |x2 + 3x| + C

Evaluate the following:

3.2(x 2) dxx 1+

+ò 4.6logx 5logx

4logx 3logx

e e dxe e

--ò 5.

(1 cos x) dxx sinx++ò

6.dx dx

1 cosx+ò 7. 2 4tan x sec x dxò 8.sinx cosx dx

1 sin2x+

+ò

9. 1 sinx dx+ò 10.x dx

x 1+ò 11.a xa x

+-ò

12.

12

34

x dx1 x+

ò 13.2

4

1 x dxx+

ò 14. 2

dx dx16 9x-

ò

15. 2

dt

3t 2t-ò 16. 2

3x 1 dxx 9

-

+ò 17. 25 2x x dx- +ò

18. 4

x dxx 1-ò 19.

2

4

x dx1 x-ò put x2 = t 20. 22ax x dx-ò

21.1

32 2

sin x dx(1 x )

-

-ò 22.

(cos5x cos4x) dx1 2cos3x

+-ò 23.

6 6

2 2

sin x cos x dxsin xcos x

+ò


EXERCISE # 1

PART # I

A-1. (A) A-2. (A) A-3. (B) A-4. (A) B-1. (B) B-2. (C) B-3. (A)B-4. (D) B-5. (D) B-6. (C) B-7. (B) B-8. (C) B-9*. (B , C) C-1. (C)C-2. (A) C-3. (C) C-4*. (C, D) D-1. (B) D-2. (D) D-3. (A) D-4. (A)D-5. (C) E-1. (B) E-2. (C) E-3. (B) F-1. (C) F-2*. (A, C, D)

PART # II

A-1. (i) 6(2x 3) C

12+

+ (ii) cos2x C

2- + (iii)

tan(4x 5) C4

++

(iv) 1 n | sec(3x 2) tan(3x 2) C |3

+ + + +l (v) 1 n | sec(2x 1) | C2

+ +l (vi) 3x 42 C

3 n2

+

+l

(vii) 1 n | 2x 1| C2

+ +l (viii) 4x 5e C4

+

+ (ix) 3/ 22(x 1) C

3+

+ (x) 2x 1 C+ +

A-2. (i) cos2x n | x 1| C

2- + + +l (ii) 4x 51 1n | sec(3x 1) | e C

3 4++ + +l

(iii) 1 n | sec(4x 5) | C2

+ +l (iv) x 1 sin2x C2 4

- +

(v) x 1 sin2x C2 4

+ + (vi) 1 1cos5x cos x C

10 2- + +

(vii) ( )3/ 2 3 / 22 (x 3) (x 2) C3

+ + + +

B-1. (i) 21 cos x C2

- + (ii) 21 n | x 1| C2

+ +l (iii) 21(tanx) C2

+ or 2sec x C

2+

(iv) xn | e x | C+ +l (v) n | x cos x | C+ +l (vi) 2x1 n | e 2 | C2

- +l (vii) 21 n | x sin2x 2x | C2

+ + +l

(viii) n | n(sec x tanx) | C+ +l l (ix) 3/ 2 1/ 22 (x 2) 4(x 2) C3

+ - + +

(x) 2x 2x1(e e ) 2x C2

-- + + (xi) 3x 2x x1 e e e C3

+ + +

(xii) 5

1 1n 1 C5 x

- + +l


C-1 (i) sin x – x cos x + c (ii) 2 2x xn x C

2 4- +l (iii)

2x x 1sin2x cos2x C4 4 8

- - +

(iv) 2

1 1x x 1tan x tan x C4 2 2

- -- + + (v) x( nx 1) C- +l

(vi) sec x tanx 1 n | sec x tanx | C

2 2+ + +l (vii) 22 x(x 1) e C- +

(viii) 1 1x 1 x 1xsin x sin x C2 2

- --+ - + (ix)

1 21 21 (tan x)x tan x n (1 x ) C

2 2

-- - + - +l

(x) xe (sinx cosx) C

2- + (xi) ex tan x + C

D-1. (i) 2 2x x 4 2 n x x 4 C2

+ + + + +l (ii) 11 xtan C2 2

- +

(iii) 2n x x 4 C+ - +l (iv) 11 xtan C

5 5- +

(v) 2 2x 1 x 2x 5 2 n x 1 x 2x 5 C2+

+ + + + + + + +l (vi) 11 (x 1)tan C

2 2- +æ ö +ç ÷

è ø

(vii) 2 3 / 2

2 1(1 x x ) 3 15 2x 1(2x 1) 1 x x sin C3 8 8 5

- æ ö- - +- - + - - - +ç ÷

è ø

(viii) 2 14 2x 3n | x 3x 4 | tan C

7 7- +

+ + - +l (ix) 3

3 3 5 / 2 3 3 3 / 22 2a(a x ) (a x ) C15 9

+ - + +

(x) 4/ 5

5

1 11 C4 x

æ ö- + +ç ÷è ø(xi)

2 3 / 2

3

(x 8) C24x-

+

(xii) x – arctan x + 21 xn C

x+

+l

D-2. (i) x 1n Cx 2

++

+l (ii) 2 11 1 3n | x 3 | n | x 1| tan x C

10 20 10-+ - + + +l l

(iii) 1n | x 1| n | x 2 | C

(x 1)- + - + + +

+l l (iv)

1 1n | x 1| n | x 2 | n | x 3 | C2 2

+ - + + + + +l l l

E-1 (i) 12 tanx / 2tan C

3 3- æ ö

+ç ÷è ø

(ii) 12 xtan 3 tan C

23- æ ö +ç ÷

è ø

(iii) 10 2x n | 3cosx 2sinx | C13 13

- + +l (iv) xn 1 tan C2

+ +l


(v) 11 3 tanxtan C

6 2- æ ö

+ç ÷ç ÷è ø

(vi) n |1 cot x | C- +l

(vii) 1 3xtanx sin2x C4 2

+ - +

F-1. (i) 2

1

1x 11 x 1 1 xtan n C142 3 3x x 1x

-+ -æ ö-

- +ç ÷è ø + +

l (ii) 2

11 x 1tan C2 2 x

- æ ö-+ç ÷ç ÷

è ø

(iii)

1x 31 xn C12 3 x 3x

+ -- +

+ +l

F-2. (i) x 2 1n Cx 2 1

+ -+

+ +l

(ii) 11 t 3 1n tan (t) C,24 3 t 3

--- +

+l where t = x 1+

(iii) 21 1 1 2n t t C,

3 3 93æ ö æ ö- - + - + +ç ÷ ç ÷è ø è ø

l where t = 1

x 1+

(iv) 2

12

x 2tan Cx

- +- +

PART # III

1. (A ® p), (B ® p), (C ® r), (D ® s)2. (A ® s), (B ® q), (C ® r)3. (A) 4. (A) 5. (B) 6. (A) 7. (B) 8. (A) 9. (C)10. (A) 11. (A)

EXERCISE # 2

PART # I

1. (B) 2. (B) 3. (C) 4. (D) 5. (C) 6. (D) 7. (A)8. (C) 9. (A) 10. (B) 11. (A) 12. (B) 13. (C) 14. (A)15. (C) 16. (D) 17. (A) 18. (A, B) 19. (B, C, D ) 20. (A , B, C, D)21. (A, B, D) 22. (B , D) 23. (A, C, D) 24. (A, B)


PART # II

1.1 sin(x a)n C

cos(a b) cos(x b)-

+- -

l 2.1 1n | sec x | n | sec 2x | n | sec 3x | C2 3

é ù- - + +ê úë ûl l l

3.3 / 2

13 / 2

2 xsin C3 a

- æ ö+ç ÷

è ø4.

22 1 4 4

2

1 x 1a sin a x C2 2a

- æ ö+ - +ç ÷

è ø

5. arcsecx – 2

nx Cx 1

+-

l

6. (a + x) arc tanx ax Ca

- +

7. esinx (x–secx) + C 8. 1(x 1) 2 x 1 2 n | x 2 | 2 tan x 1 C-+ + + - + - + +l

9. 2 12 n sin 4sin 5 7 tan (sin 2) C-f - f + + f - +l 10.( ) ( )3 / 22 2

5

4 x x 6C

120x

+ -+

11.1 13x sin4x sin8x C

128 8é ù- + +ê úë û

12. ( )11 1tan 2 tanx tanx C22 2

- + +

13.2 2

1 11 x 1 2 2x 1tan tan C3 x 3 3 3

- -æ ö æ ö- +- +ç ÷ ç ÷

è ø è ø14. sin x 2 2sin x1n (xe ) n (1 x e ) C

2- - +l l

EXERCISE # 3

PART # I

1. 1 22

3 1 1 xtan x n (1 x) n(1 x ) C2 2 4 1 x

- - + + + + ++

l l

2. (C)

3. ( )1 22x 2 3(x 1) tan n 4x 8x 13 C3 4

- +æ ö+ - + + +ç ÷è ø

l

4.( )

m 13m 2m m m2x 3x 6x

C6 (m 1)

+

+ ++

+5. (D) 6. (A) 7. (D) 8. (C)

PART # II

1. (A) 2. (B) 3. (D) 4. (A) 5. (A) 6. (C) 7. (D)


EXERCISE # 4

indefinite integration - arride learning€¦ · topic p n. teory 01 - 03 ecis - 1 04 - 11 ecis - 2...

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