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o SOLUTIONS OF * MARKED PROBLEMS (INTEGRAL CALCULUS)

1

Miscellaneous Exercises (Integral Calculus) (Solutions of * marked problems) 1. Let, I =

=

z

z

x + 1 dx = ( x + 2) x + 3

Now, I1 =

z

=–

z z

z

2 z dz

•taking x + 1 = z2 or,, dx = 2z dz—

z z +2 2

z2 + 2 | = 2log [ x + 1 + x + 3 ];

dx •taking x + 2 = 1 or, dx = – 12 dt— t ( x + 2 ) ( x + 1)( x + 3) t

1 t

dt t2

e je j 1−1 1+1 t t

=–

Þ I = log[x + 2 +

z

dt t 1− t2 t2

=–

z

dt = – sin–1t = – sin–1 1 . 2 x+2 1− t

1 x + 1 + x + 3 + sin–1 x + 2

\ I = I1 – I2 = 2 log

2.

dx ( x + 2 ) ( x + 1)( x + 3) = I1 – I2 •say—.

dx = ( x + 1)( x + 3)

= 2 log | z + I2 =

z

dx – ( x + 1)( x + 3)

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

z

x2 + 4 x + 3 ] + sin–1

1 + c. x+2

z

dx (ax + b ) + ax + b

=

z

=

2 dz 2 2 = log(z + 1) + c = log ax + b + 1 + c . a z +1 a a

2 zdz •taking z = a( z2 + z)

ax + b , we get 2z dz = adx —

z

3. Hints : I =

e

1 2

z

( x2 + 1) − ( x2 − 1) 1+ x4

dx=

j

1 2

z

1 + 12 x

x2 + 12 x

dx –

1 2

z =

1

1 − 12 x

x2 − 12

dx.

x

1 2

ze

d x − 1x

e

j

2 x − 1 +2 x

j

–

1 2

ze

d x + 1x

e

j

2 x + 1 −2 x

j

.

2

RUDIMENTS OF MATHEMATICS

8.

2 −1 −1 −1 Let, I = etan x . 1 + x + x dx = etan x dx + etan x x 2 dx = I1 + I2. 1+ x 1 + x2

FG H

z

IJ K

z

z

z z LMN

−1 −1 I1 = etan x dx = etan x 1dx −

z

= xetan

−1 x

z

− etan

−1 x

\ I = I1 + I2 = x e tan

.

IJ z OP K Q

FG H

d tan −1 x e . 1 dx dx + c dx

1 tan −1 x . xdx + c = x e – I2 + c 1 + x2

−1 x

+ I2 – I2 + c = x e tan

−1 x

+ c.

e xx−1 j dx e x +x3x +1 j tan−1e x x+1 j 2

9.

Suppose, I =

ze

( x − 1) dx = 2 x + 3 x + 1 tan−1 x x+1 2

4

2

j

e j

z

2

4

2

2

2

1− 1 j 1− 1 e x x = dx = 2O 2 1 − 1 1 L 1 e x + 3 + x j tan e x + x j MN1 + e x + x j PQ tan−1e x + 1x j O 1 = dt + c LM let, tan–1 e x + 1x j = z or,, . 1 − 1 j dx = dz P 2 e x z Q N 1 + e x + 1x j = log | z | + c = log tan−1e x + 1x j + c.

z

z

2

2

2

z

10.

z

2

sin x dx = sin 4 x

= 1 4

z z LM Nz

z

sin x dx = 2 sin 2 x cos 2 x

z

sin x dx 4 sin x cos x cos 2 x

z

dx dx = 1 4 cos x(1 − 2 sin2 x) cos x cos 2 x

z

LM suppose, sin x = z or, cos x dx = dzOP Q N

z

cos xdx dz dz = 1 = 1 = 1 2 (2 − 2 z2)(1 − 2 z2 ) 4 cos2 x(1 − 2 sin2 x ) 4 (1 − z2 )(1 − 2 z2 ) = 1 2 =

OP Q

LM N

1 1 1 + 2z 1 1+ z dz − dz log = − log 2 2 2 2 2 1− z . 2 2 1 − 2z 1 − 2z 2 (1 − z )

z

1 log z − 1 – 1 log 2 z − 1 + c 8 z+1 4 2 2z + 1

OP + c Q

o SOLUTIONS OF * MARKED PROBLEMS (INTEGRAL CALCULUS)

= 1 log sin x − 1 – 1 log 2 sin x − 1 + c. 8 sin x + 1 4 2 2 sin x + 1 11.

I=

z

sec x dx = 2 sin( x + a ) cos x = =

13.

z =

=

1 2

1 2

3

z

sec2 x dx sin x cosa + cos x sina

sec2 x dx = cosa ⋅ tan x + sina

z

2 2 cosa

z

1 2 cosa

cosa tan x + sina =

d (cosa tan x + sina ) cosa ⋅ tan x + sina

2 tan x ⋅ seca + tana seca . (Ans.)

dx ( x + 1) x2 + 6 x + 7

z z

LM Suppose, x + 1 = 1 , then dx = − 12 dz. OP z Q z N

dx ( x + 1) ( x + 3)2 − 2 − 12 dz

= −

z

1 z

2

e 1z + 2j

−2

1 2

z

dz ( z + 1)2 −

FH 1 IK 2 2

2 = − 1 log z + 1 + (z + 1)2 − 1 +c 2 2

e j

e x1+1 j2 + ( x2+1) + 12

= − 1 log 1 + 1 + x +1 2

+c

2 = 1 log | x + 1 | – 1 log x + 2 + x + 6 x + 7 + c. 2 2 2

14.

z

x2dx = ( x sin x + cosx )2

=

x cos x

=–

z

z

x cosx . x dx 2 cos x ( x sin x + cos x)

d ( x sin x + cos x ) – ( x sin x + cos x )2

x . 1 + cosx ( x sin x + cos x)

z LMN z

FH

d x dx cos x

IK

z

OP Q

d ( x sin x + cosx ) dx + c ( x sin x + cosx )2

1 . (cosx + 2x sin x) dx + c ( x sin x + cos x) cos x

3

4

RUDIMENTS OF MATHEMATICS

x = – cosx ( x sin x + cos x) + sec2 x dx + c

z

x x + tanx + c = – + sin x + c cosx ( x sin x + cos x) cosx ( x sin x + cos x) cosx

=–

2 2 = − x + x sin x + sin x cos x + c = sin x cosx − x cos x + c cos x( x sin x + cosx ) cosx ( x sin x + cos x)

= 15.

cosx (sin x − x cos x) + c = sin x − x cos x + c. cosx ( x sin x + cos x) x sin x + cos x

z

1

Hints : I = (x3m–1 + x2m–1 + xm–1 ) ( 2 x 3m + 3 x 2 m + 6 xm ) m dx. Put 2x3m + 3x2m + 6xm = z, then dz = 6m (x3m–1 + x2m–1 + xm–1 )dx 1 \ I= 6m

16.

z

m +1

m+ 1 1 z m 1 3m 2m m m z dz = = ( 2 x + 3 x + 6 x ) . 6 m m+1 6 ( m + 1) 1 m

m

dx = x +3 x

=

3 z5dz 6 z5dz = 6 = 6 z + 1 − 1 dz 3 2 2 z +1 z +z z ( z + 1)

z

z

LM suppose,, x N

z

dx 1 x + x3 1 2

z

1 6

= z or, x = z6 or, dx = 6z5 dz

OP Q

z

LM ( z + 1)(z2 − z + 1) dz − dz OP = 6 z (z2 − z + 1)dz − log | z + 1 | z + 1Q N (z + 1) L z3 z2 O = 6 M − + z − log | z + 1|P + c = 6 L 1 x − 1 x + x − log x + 1 O + c MN 3 2 PQ N3 2 Q

z

=6

z

1 2

1 3

1 6

1 6

= 2 x – 3 3 x + 6 6 x – 6log 6 x + 1 + c. 17.

z =

dx = x +1 − 3 x +1

z

=6

zb

g b g

x +1 − x +1

1 3

LM suppose,, b x + 1g = z or, (x + 1) = z6 or, dx = 6z5 dz OP Q N z5 dz = 6 L z3 − 1 dx + dz O + c MN z − 1 z − 1 PQ z2 ( z − 1)

6 z5dz z3 − z2

z

dx 1 2

1 6

z

z

o SOLUTIONS OF * MARKED PROBLEMS (INTEGRAL CALCULUS)

=6

LM N

5

OP Q

z3 + z2 + z + log | z − 1| + c ( z2 + z + 1)dz + log | z − 1| + c = 6 3 2

z

1

b g

1

b g

1

1

= 2 x + 1 2 + 3 x + 1 3 + 6 x + 1 6 + 6 log x + 1 6 − 1 + c.

b g

18.

z

1 1 − x dx = x 1+ x

z

=–4

z

1 − cos2q (–4 cos 2q sin 2q ) dq 1 + cos2q

1 cos2 2q

LM suppose,,þ N

b g

x = cos 2q or,, x = cos2 2 q or, dx = – 4 cos 2q sin 2q d q

2 1 . tan q . sin 2q d q = – 4 2 sin q d q = – 4 1 − cos 2q d q cos 2q cos 2q cos2q

z

z

= – 4[ sec 2q d q – d q ] = – 2 log | sec 2q + tan 2q | + 4q + c

z

= – 2 log

19.

z

1 + sin 2q + 2 cos–1 x + c = 2 cos–1 x – 2 log 1 + 1 − x + c. cos2q x

F GH

x 2 + 1 = z or, 1 +

Suppose, x +

FH

IK

or, x + x2 + 1 dx =

Again, x +

I JK

1.2 x dx = dz 2 x2 + 1

x 2 + 1 dz or,, dx =

FH x +

x2 + 1

IK dz.

x2 + 1

x 2 + 1 = z Þ x2 + 1 = (z – x)2

2 Þ x2 + 1 = z2 – 2zx + x2 or, x = z − 1 . 2z

Then,

=

z

z

x + x2 + 1 dx =

2 z − z2−z1 dz = z

z

z

x2 + 1

dz

x + x2 + 1

2 z2 − z2 + 1 dz = 1 z2 + 1 dz = 1 z12 dz + 1 z− 32 dz 3 2 2 2 2z z z2

z

z

−1 2 32 1 = z + z 2 + c = 1 . x + x2 + 1 3 2. 3 2. − 12

e j

FH

IK

3 2

−

z

1 x + x2 + 1

+ c.

OP Q

6

RUDIMENTS OF MATHEMATICS

20.

z

dx

=

3 4

z

1 3 4

dx

x e x + 1j x 2. x 3 . F 1 + 1 I H xK LM suppose,, 1 + 14 = z, then – 45 dx = dz or,, dx = – x5 dz. OP 4 Q x x N F I = − 1 dz = – 11 z + c = – G1 + 14 J + c = – 1 e1 + x4 j + c. x 4 z H xK 4. 4 2

4

4

z

21.

3 4

Hints : I =

1 4

1 4

z

2

F H

I K

1

1 4

1

x3 x3 + 1 + 2x6

F I x 1+ x H K 1 3

dx =

z

x

− 13

dx + 2

z

dx

x

5 6

FG 1 + x IJ H K 1 3

and then in the 2nd integral put

1

z = x6 .

22.

zd

dx = 1 + x x − x2

i

zb

2 sin q cosq d q 1 + sinq

g

sin 2 q − sin4 q

LM suppose, x = sin2 q . \ dx = 2sin q cos q dq OP Q N =2

=2

zb z

sinq cosq d q 1 + sinq

g

sin 2 q (1 − sin2 q )

=2

sin q cosq d q 1+ sin q sin q cosq

zb

g

1 − sin q 1 − sin 2q dq dq = 2 dq =2 2 1 + sin q cos2 q 1 − sin q

z

z

= 2 sec2 q dq − secq tanq dq = 2(tan q – sec q ) + c

z

z

FH sinq − 1 IK + c = 2 FH sin q − 1 IK + c = 2 FG cosq cosq cos q H F I = 2 G x − 1 J + c. H 1− x K

=2

23.

z

cos−1 1 − x 3 dx = 2

z

I +c J 1 − sin q K sinq − 1 2

cos−1 1 − sin2 q . 3 sin2 q cosq d q

o SOLUTIONS OF * MARKED PROBLEMS (INTEGRAL CALCULUS)

7

LM suppose, x = sin3q ; then, dx = 3 sin2 q cos q dq OP Q N = cos–1(cosq ). 3 sin2q cosq dq

z

z RST ddq (q )z sin q cos q dq UVWd q OPQ LF I O L O = 3 Mq G sin q J − z sin q dq P = 3 Mq sin q − 1 z (3 sin q − sin 3q )d q P N H 3 K 3 Q N 3 12 Q = 3[q

z sin

2

q cosq d q –

3

2

3

3

= q sin3 q – 3 (– cos q ) – 1 cos 3q + c 12 4 = q sin3 q + 3 cos q – 1 (4cos3 q – 3 cos q ) + c 4 12 = q sin3 q + cos q – 1 cos3 q + c 3

1

FH

2

2

= x sin–1 x 3 + 1 − x 3 − 1 1 − x 3 3 24.

=

3 2

+ c.

x2 + 1 {log(x2 + 1) – 2 log x} dx x4

z =

IK

z z

x 2 +1 x2 3

x

{log(x2 + 1) – log x2}dx =

1 + 12 x

x

3

FG H

IJ K

log 1 + 1 dx = – 1 2 x2

z

z

1 + 12 x

x

3

FG H

IJ K

2 log x 2+ 1 dx x

z log z dz

LM suppose,, FG1 + 1 IJ = z; then – 23 dx = dz or, – dx3 = dz OP 2Q H x2 K x x N O 1 L2 1 2 = − 1 Llog z. 2 z − 2 z dzO = M z − log z.z P + c = z LM − log zOP + c 3 3 2 NM 3 3 N Q QP 3 N 3 Q 1F F 1 IO 1 I L2 = G 1 + 2 J M − logG 1 + 2 J P + c. 3H H x KQ x K N3 3 2

3 2

z

1 2

3 2

3 2

3 2

8

RUDIMENTS OF MATHEMATICS

25.

z

x + 1 + 1 − x dx

d

i

log

= log

d

= x log

= x log

= x log

= x log

= x log

= x log 26.

z

x + 1 + 1 − x dx –

i

d d

x +1 + 1− x –

i

i

2

d d

x +1 + 1− x + 1 2

d d

d

x +1 + 1− x

x x +1 + 1− x

1 x +1 + 1− x + 2

i i

z z

e

F iH2

x . x +1 + 1− x

1+ x − 1− x

I K

d

1− x − x +1 1 − x2

x 1 − x2

1 − x2

2 x. 1 − x2

z

z

1 1 − dx x +1 2 1− x

j2 .

FH x + 1 + 1 − x − 2

i

it dxOPQ dx

i

x +1−1+ x

x + 1 + 1− x + 2 .1 2 2

FH1 −

1 − x2 1 − x2

i dx

dx

IK . x dx

IK dx

x + 1 + 1 − x + 1 sin–1x – x + c.

i

2

2

z z LMN

dx log(1 − x) dx = log(1 – x) 2 − 2 x x

= – 1 log(1 – x) + x

z

d log dx

x +1 + 1− x – 1

= – 1 log (1 – x) – x

27.

z LMN o zd zd

z z FH

FH

q dxx OPQ dx

d log(1 − x ) dx

l

z

2

F I 1 1 . dx H K x −1 x x −1 + c. − 1 I dx = – 1 log (1 – x) + log K x

(−1) . − 1 dx = – 1 log (1 – x) + (1 − x) x x 1 x −1

x

z

x

IK

cos2q log cosq + sin q d q cosq − sin q

= log

F cosq + sinq I H cosq − sinq K

z

cos2q d q −

z LMN

RS F T H

cos q + sin q d log cos q − sin q dq

I UV KW

z

OP Q

cos 2q dq d q

o SOLUTIONS OF * MARKED PROBLEMS (INTEGRAL CALCULUS)

=

F H

sin 2q log cosq + sinq 2 cosq − sinq

I K

–

z FH

cosq − sin q cosq + sin q

I K

×

RS (cosq − sinq )(cosq − sinq ) − (cosq + sinq )( − cosq − sinq ) UV. sin 2q |T |W 2 (cosq − sin q ) sin 2q logF cosq + sinq I cosq − sin q I {2 cos q + 2 sin q } sin 2q F = – H cosq − sinq K z H cosq + sinq K . (cosq − sinq ) . 2 dq 2 sin 2q . 2 sin 2q logF cosq + sin 2q I = dq H K–z 2

2

2

2

2

cosq − sin q

2(cos2 q − sin2 q )

F H

I K

1 sin 2q cosq + sin q = 2 sin 2q log cosq − sin q − cos2q

z

FH cosq + sinq IK – z tan 2q dq cosq − sinq = 1 sin 2q log F cosq + sinq I – 1 log | sec 2q | + c 2 H K 2 = 1 sin 2q log 2

cosq − sinq

= 1 log | cos 2q | + 1 sin 2q log 2

28.

z

2

dx = secx + cosecx

z

1 cos x

dx = + sin1 x

FH cosq + sinq IK + c. cosq − sinq

z

z

sin x cos xdx = 1 1 + sin 2 x − 1 dx 2 sin x + cos x sin x + cos x

LM b OP g2 MN PQ L OP 1 1 dx = M (sin x + cos x)dx − 2M 2 sine p + x j P 4 N Q 1 1 cosecF x + x I = b− cosx + sin xg − H 4 K dx 2 2 2 1 sin x + cos x dx = 2 (sin x + cos x ) dx − sin x + cos x

z

z

z

z

= 1 (sin x – cos x) – 2

1 2 2

log tan

FH x + p IK 2 8

9

+ c.

dq

10

RUDIMENTS OF MATHEMATICS

29.

z = = =

=

=

30.

dx = sin x + sec x

= =

z

dx = 1 sin x + cos x

cos xdx = sin x cos x + 1

z

2 cos xdx 2 + 2 sin x cos x

(cos x + sin x ) + (cos x − sin x) dx 2 + 2 sin x cos x

z z z zd

cos x + sin x dx + 2 + 2 sin x cos x

z

cos x − sin x dx 2 + 2 sin x cos x

sin x + cosx dx + 3 − (sin x − cos x)2

z

d (sin x − cosx )

+

2

3 − (sin x − cos x )

i

2

cos x − sin x dx 1 + (sin x + cosx )2

z

d (sin x + cos x ) 1 + (sin x + cos x )2

1 log 3 + sin x − cosx + tan–1 (sinx + cos x) + c. 2 3 3 − sin x + cosx

In = =

z

z z z

z

sin nx dx = sin x

z

sin( n − 2 + 2 ) x dx sin x

sin( n − 2 ) x cos2 x + cos(n − 2) x sin 2 x dx sin x sin(n − 2) x(1 − 2 sin2 x) dx + sin x

z

cos(n − 2) x sin 2 x dx sin x

sin( n − 2 ) x – 2 sin(n – 2)x sin x dx + 2 cos(n – 2)x cos x dx sin x

z

z

z + 2 z cos(n – 2 + 1)x dx = I

= In–2 + 2 [cos(n–2)x cosx – sin(n – 2)x sin x]dx = In–2

n–2

or, In – In–2 = 31.

+

2 sin (n – 1)x + c n−1

2 sin(n – 1)x + c or, (n – 1)(In – In–2) = 2 sin (n – 1)x + c. n−1

z

z

In = secn x dx = secn–2 x. sec2 x dx = secn–2 x

z sec

2

x dx –

z LMN

d secn−2 x dx

e

z sec

= secn–2 x tan x – (n – 2)

z

n–3

j

z

OP Q

sec2 x dx dx

x sec x tan x. tan x dx

= secn–2 x tan x – (n – 2) secn–2 x (sec2 x – 1)dx

o SOLUTIONS OF * MARKED PROBLEMS (INTEGRAL CALCULUS)

= secn–2 x tan x – (n – 2)

z sec

n

z sec

x dx + (n – 2)

n–2

11

x dx

or, In + (n – 2) In = secn–2 x tan x + (n – 2) In–2 + c n− 2 or, In = sec x tan x + n − 2 In–2 + c. n −1 n −1 2

32.

FH

z

1 2

IK

z

2

2 1 2

=–

2

FH IK

z

z

1 sin 1 − t dt = t t

1 2

2

\ 2I = 0. I=

FH IK

1 sin 1 − t dt = – t t

2

z 1 2

F I dt = – I. H K

1 1 sin t − t t

\ I = 0.

LM MN

OP PQ

p b b p x cos2 ( − x ) cos2 x dx = a ⋅ cos2 x dx. dx by f ( x ) dx = f ( a + b − x ) dx = x x 1 + a− x −p 1 + a a a −p 1 + a −p p

33.

FH IK FG JI LM H K N

Suppose, I2 = 1 sin x − 1 dx = t sin 1 − t . − 12 dt let 1 = t; then, – 12 dx = dt t x x x t x 1

z

z

\ 2I =

=

p

z

(1 + a x ) cos2 x

−p

1 + ax

1 2

LM x + sin 2x OPp N 2 Q−p

z

z

p

dx =

z

cos2 x dx =

−p

=

1 2

z

p

z

(1 + cos 2x) dx

−p

1 .2p = p. 2

\ I = p (Ans.) 2

34.

Suppose, I =

np + v

z

| sin x | dx =

np

z

0

| sin x | dx +

np + v

z

| sin x | dx.

np

0

Since, | sin(p + x) | = | – sin x | = | sin x |, np + v

p

I=n

z | sin x | dx + I , where I 1

0

p

=n

z sin x dx + I

1

1

=

z

| sin x | dx

np

= n [ − cos x ]p0 + I1 = n[1 + 1] + I1 = 2n + I1.

0

Let, x = n p + q or, dx = d q . when, x = n p , q = 0; when x = n p + v, q = v.

OP Q

12

RUDIMENTS OF MATHEMATICS np + v

Then, I1 =

z

np

| sin x | dx =

v

z sinq dq = b− cosq g

v 0

0

= 1 – cos v.

\ I = 2n + I1 = 2n + (1 – cos v) = 2n + 1 – cos v. 35.

z

x

loge x2

loge x x dx =

z

xlog e x

2

loge x loge e

LM suppose,, y = xlog x N e

dx = x

2

z

2

xloge x .

loge x dx x

or, log g e y = loge x2 loge x or, loge y = 2loge x loge x or, loge y = 2(loge x)2

or,

1 dy = 2.2(log g e x). 1 y dx x

2 \ dy = y.4(loge x). 1 dx = xloge x . 4 loge x dx. x

x

= 1

37.

z

0

z

OP Q

2 1 dy = 1 y + c = 1 xloge x + c. 4 4 4

dx = 2 ( x − 2 x + 2)2

1

z

0

dx = [( x − 1)2 + 1]2

z

dx [( x − 1)2 + 1]2

LM suppose, x – 1 = tan q; then, dx = sec2 q dq OP Q N =

z

sec2q d q = cos2 q d q = 1 (1 + cos 2q ) d q 4 2 sec q

z

z

OP = 12 LM tan−1( x − 1) + x − 1 2 OP . Q N 1 + ( x − 1) Q 1 1 1 L tan−1( x − 1) + x −1 O dx \ = 2M P 2 1 + ( x − 1)2 Q0 N − 2 x + 2 )2 ( x 0 1 −1 O 1 L −1 1 O = = LM 0 + 0 − tan−1(−1) − tan 1 + = 1 Lp + 1O = p +2 . 2N 1 + 1 PQ 2 MN 2 PQ 2 MN 4 2 PQ 8 LM N

= 1 q + sin 2q 2 2

z

39.

z

sin 2 x dx = cos2 x + 8 cos x + 7

=

z

z

2 sin x cos x dx 2 cos2 x − 1 + 8 cos x + 7

2 sin x cos x dx = 2(cos 2 x + 4 cos x + 3)

z

sin x cosx dx (cos x + 3)(cos x + 1)

o SOLUTIONS OF * MARKED PROBLEMS (INTEGRAL CALCULUS)

=–

=–

LM suppose, cos x + 1 = z, then, – sin x dx = dzOP Q N

( − sin x) cos xdx (cos x + 3)(cos x + 1)

z z

(z − 1) dz = – z(z + 2)

z z

dz + 3 dz =– z z( z + 2)

z z z

dz + 3 dz − 3 dz z 2 z 2 z+2

= – log | z | + 3 log | z | – 3 log | z + 2| 2

2

= 1 log | z | – 3 log | z + 2 | = 1 log | cos x + 1| – 3 log | cos x + 3 | 2

2

p 2

z

∴

0

2

2

p sin 2 x dx = 1 log cos x + 1 − 3 log cos x + 3 2 cos2 x + 8 cosx + 7 2 0

= 1 [log1 – 3 log3 – log2 + 3 log4] 2

4 = – 1 log2 + 3 log . 2 2 3 p

40.

Let, I =

z x sin 2x. sin( p2 cos) dx

0

p

= (p – x) sin(2p – 2x) sin[ p cos(p – x)] dx

z

2

0

p = (p – x) ( – sin 2x) sin[– p cos x] dx

z

2

0

p

p

= p sin 2x sin( p cos x)dx – x sin 2x sin( p cos x) dx

z

2

0

=p

z

0

2

p

z 2 sin x cos x sin ( p2 cos x) dx – I.

0

So, 2I = p

13

p

z 2 sin x cos x sin( p2 cos x) dx

... (1)

0

z 2 sin x cos x sin( p2 cos x) dx = – 2 z z sin( p z) dz •suppose, cos x = z or, – sin dx = dz 2

or, sin x dx = – dz—

14

RUDIMENTS OF MATHEMATICS

LM FH IK LM d ( z) sinFH p zIK dzOP dzOP 2 N N dz Q Q L 2z F p I 2 F p I O = – 2 M − p cos H 2 zK + p z cos H 2 z K dz P N Q L pz pz O = – 2 M − 2 z cosF I + 4 sin F I P . N p H 2 K p H 2 KQ = – 2 − z. 2 cos p z – 2 p

z z

2

\ from (1) we get p L O 2 cos x p 4 p F I F I cos cos sin cos x x 2I = – 2p M − N p H 2 K + p 2 H 2 K PQ0 L = – 2p M − 2 cosp cosF p cosp I + 42 sin F p cosp I N p H2 K p H2 K

FH

IK L2 p 4 p O \ I = – p M cosFH − IK + 2 sinFH − IK + 2 cos0 cos p − 42 sin p P 2 2 p 2 2 p N p p Q = – p L0 − 4 + 0 − 4 O MN p 2 p 2 PQ F I = – p G − 82 J = 8p2 = 8 . H pK p p

FH

+ 2 cos0 cos p cos0 − 42 sin p cos0 2 2 p p

41.

Let, I =

z

= tan–1

tan−1

x − 1 dx

x − 1 dx –

z

z LMN

zd z

= x tan–1

x −1 –

= x tan–1

x −1 −

z

LM suppose, N

dx x −1

1 4

d tan−1 dx

{

1 . 1+ x −1 2

i

1 x −1

} dxOPQdx

x −1

z

1 . 1 x dx x −1 2 x

dx ... (1)

x = z, x = z2; then dx = 2z dz

OP Q

IK OP . Q

o SOLUTIONS OF * MARKED PROBLEMS (INTEGRAL CALCULUS)

=

2 z dz = 2 z − 1 + 1 dz = 2 z −1 z −1

z

z

= 2. 2 z − 1

b g 3

3 2

b g

+ 2. 2 z − 1

x −1 − 1 .4 4 3

\ I = x tan–1

x −1 − 1 3

= x tan–1

\

16

z

tan−1

I=

z 0

4 3

x −1

d

i

x −1

d

i

3

3 2

3 2

1 dz z−1

e

d

1

x −1 2.

i

x −1

j

1 2

1

LM N

x −1 −

i

z

+4

– 1⋅4 4

x −1 2.

i d

x − 1 dx = x tan−1

1 3

d

3

x −1 2 −

i d

16

i OPQ

x −1

1 2

1

3 1 1 1 p 3 − 1 .32 − 32 = 16 p − 3 2 − 32 = 16 − 2 3 . 3 3 3

= 16 tan–1

42.

=

z − 1 dz + 2

x −1 2 −

d

1

p 2

1 2

z

p 2

sec x dx (1 + tan2 x )sec2 x = dx. 2 2 2 2 2 2 2 (a + b tan x ) 0 ( a + b tan x ) 4

z

2

Put b 2 tan2 x = a 2 tan2 q. When x = 0; then q = 0. When x = p ; then q = p . 2 2

FH 1 + a b \ I= z p 2

2 2

0

tan2 q

IK ab sec2 q dq =

a4 sec4 q

p 2

1 1 (b 2 cos2 q + a 2 sin 2q ) dq = 3 3 I1 , 3 3 a b 0 a b

z

p 2

where I1 =

z

(b 2 cos2 q + a 2 sin2 q ) dq

0 p 2

=

z RST 0

FH

IK

FH

b2 cos2 p − q + a2 sin2 p − q 2 2

IK UV dq = z (b2 sin2q + a2cos2q ) dq. W 0

p 2

\ 2I1 =

z

p 2

(a 2 + b 2 ) (sin2 q + cos2 q )dq = (a 2 + b 2 )

0

\ I=

p 2

( a 2 + b2 ) p . (Proved.) .) 4 a3b 3

z 0

dq = (a 2 + b 2 ) p . 2

15

16

RUDIMENTS OF MATHEMATICS

1 x

43.

log t Here f F 1 I = z H x K (1 + t ) dt

1 1 , then dt = – 2 dz. z z

Put, t =

1

F I = zx log 11z FG − 12 IJ dz = zx log z dz = zx logt H x K 1 1 + z H z K 1 z(1 + z) 1 (1 + t )t

\ f 1

t

1

1/x

z

1

x

dt.

F I = zx LM 1 + 1 OP log t dt = zx logt dt. H x K 1 N 1 + t t (1 + t) Q t 1

\ f(x) + f 1 x

Now I =

z

1

logt x dt = log t ⋅ log t 1 – t

\ 2I = (log x)2

F I = (log x )2 H xK 2

1

logt dt = (log g x)2 – I t

(log x ) 2 . 2

or, I =

\ f(x) + f 1

x

z

(Proved.) .)

44. Put x = cos q, then

0

So I = –

z

p

sin2 q dq = a − cosq p

= (1 – a 2 )

z

0

p

z

0

dq + a − cosq

ze z 0

Now,

z

1

q

p

0

p

z

(a + cos q )dq

0

dq

p

2q 2q 2q 2q 0 a sin 2 + cos 2 − cos 2 − sin 2

p

= (1 – a 2 )

– 1

(1 − a 2 ) + ( a2 − cos2 q ) dq a − cos q

p

= (1 – a 2 )

x

j e

sec2 q2 d q

( a − 1 ) + ( a + 1) tan2 q2

sec2 q2 d q

( a − 1) + (a + 1) tan2 q2

dq =

2 a +1

j

+ aq + sinq 0

+ ap.

ze

dz a−1 a+1

2

j

+ z2

(putting z = tan

q so that dz = 1 sec2 q dq ) 2 2 2

o SOLUTIONS OF * MARKED PROBLEMS (INTEGRAL CALCULUS)

2

=

( a2

− 1)

. tan–1

FG H

\ I = ap – (a 2 – 1)

a + 1z a −1

IJ = K

LM MN

2

lim

q →p

2 a −1

q →p

a2 − 1

2

a −1

tan

FH

LM N

lim 1 + 1 + ⋅⋅⋅ + 1 4n n→∞ n n + 1 = lim 1 + lim n→∞ n

n→∞

FG H

IJ K

a + 1 tan q . a −1 2

a + 1 tan q a −1 2

FG H a2 − 1 I K

tan–1

2 = ap – 2 a − 1 p = p a − 2

51.

FG H

−1

2

= ap – (a 2 – 1) lim

tan–1

2

IJ OPq K PQ0

a + 1 tan q a −1 2

IJ K

[Q a > 1]. (Proved.) ed.)

OP Q

LM 1 + 1 + ⋅⋅⋅ 1 OP + lim LM 1 + 1 + ⋅⋅⋅ + 1 OP N n + 1 n + 2 n + n Q n→∞ N 2n + 1 2n + 2 2n + n Q + lim L 1 + 1 + ⋅⋅⋅ + 1 O 3n + n PQ n→∞ MN 3n + 1 3n + 2

n n 1 1 + lim 1 n 1 + lim 1 = 0 + lim 1 r r n 3 + rn 1 + n n→∞ n 2+ n n→∞ n→∞ n r =1 r =1 r=1

∑

∑

∑

1

z

1

1 1 1 1 dx dx dx + = + x + 3 = log( x + 1) 0 + log( x + 2 ) 0 + log( x + 3) 0 1+ x 2+ x 0 0 0

z z

= log 4. 53.

1

1 2

1

0

0

1 2

z f(x) dx = z f(x)dx + z f(x)dx

=

1 2

1 2

0

0

1 2

1 2

z f(x)dx + z f FH z + 12IK dz[in the second integral, putting z = x – 12 we get] 1 2

1 2

1 1 = z f(x)dx + z f FH x + IK dx = z {f(x) + f FH x + IK }dx = z dx = 1 . 2 2 2 0 0 0 0 1 54. I = 2

p 2

z 0

17

sin2 x dx = 1 2 1+ x

p 2

LM − 1 cos2 x OP N x + 1 2 Q0

1 – 4

p 2

z

cos 2 x dx 2 0 ( x + 1)

18

RUDIMENTS OF MATHEMATICS

LM MMN e

OP j PPQ

p 2

z

1 1 1 1 cos 2 x 1 = + – dx = 4 ( x + 1) 2 2 2 p +1 2 2 0 2 (Putting 2x = z so that dx =

=

55.

LM N

OP Q

LM 1 + 1 OP – 1 pz N p + 2 2Q 8 0

e + 1j z 2

2

dz

1 dz, and when x = 0, z = 0, and when x = p , z = p ) 2 2

FH

z

cos z

IK

1 1 +1 1 cos z 1 1 1 – dz = + − A . (Proved.) oved.) 2 p +2 2 2 ( z + 2) 2 2 2 p +2

FG H

IJ K

2 2 Here, for 0 £ x £ 2 we have, f ¢(x) = cos x + 2 x + 1 = cos ( x + 1) . 5 5 2 \ f ²(x) = – sin ( x + 1) . 2 ( x + 1) . 5 5

At a point a , in the interval 0 < x < 2, the function f(x) has a maximum value, if f ¢(a ) = 0 and f ²(a ) < 0. Now, f ¢(a ) = cos

(a + 1)2 (a + 1)2 = p =0Þ Þ 2 a 2 + 4a = 5p – 2. 5 5 2

Clearly, at that point a , f ²(a ) = − 2 sin p . 5p = − 2 5p < 0 . 5 2 2 5 2 So, at x = a , f(x) has maximum value, when 2a 2 + 4a = 5p – 2. 56.

ep

z

2

1 =– 2 1 = . 2

z + p2 j sin(p + 2 z).sine p2 cose p2 + z jj e dz z p 2

j

p x sin 2 x .sin 2 cos x I= 1 dx = 1 p 2 20 x−

z

− p2

z + p2 j sin 2 z.sine − p2 sin zj e dz z p 2

z

− p2

p 2

p 2

−p

− p2

sin 2 z sine p sin zj p F I p dz 2 sin z .sin sin z dz + z z H2 K 4 z 2

p

FH

1 2 p = .2 sin 2 z sin sin z 2 2

z

0

IK

dz

2

FH

IK

[Qf(z) = sin 2z sin p sin z is an even function and 2 sin 2 z. sin p2 sin z g(z) = is an odd function] z

e

j

o SOLUTIONS OF * MARKED PROBLEMS (INTEGRAL CALCULUS) p 2

Let, p sin z = p, then

p 2

= 2 z sin z.cosz.sin F p sin zI dz = 2 z 2 p sin p. 2 dp H 2 K 0p p 0

2 p cos z dz = dp 2

p

p 2 8 = 82 p sin p dp = 2 − p cos p + sin p 02 = 82 . p p 0 p

z

57.

F GH

58.

x

0

p

0

Here, lim f ¢(x) = lim 2 sin x −3sin 2 x = lim 2 sin x (1 3− cos x) x→0 x→ 0 x→0 x x x

sin sin x = lim x . xlim x 2 x→ 0 →0 2

2

I JK

2

= 1. (1)2 = 1.

Let f (t) be a primitive of f(t) in 0 £ t £ x, then f ¢(t) = f(t) for all t in 0 £ t £ x. x

z Now g(x) = 0 x

z

Þ x g(x) =

f ( t )dt x f(t)dt = f (x) – f (0). [by fundamental theorem of integral calculus]

0

Þ x g¢(x) + g(x) = f ¢(x) = f(x) Þ lim x g¢(x) + lim g(x) = lim f(x) x→0

Þ lim g(x) = f(0) x→0

59.

I = 41 3 = 42

3

z

z

i

z z − 4 3 dz =

•let, 4 3 + x + 3 = z. \ dx = 2(z – 4 3 )dz—

4 z52 − 16 3 z32 + c 54 3 34 3

3

[ 3 z – 20] + c = 44 4 3 + x + 3 5 27 5 27

2 = 44 z

60.

d

Hint : I =

=

z z

x→0

[Q g¢(x) exists and f is continuous att x = 0].

4 3 + x + 3 dx

d

x→0

3 2

i

3 x + 9 − 8 + c.

(sin2 q − cos2 q ) (sinq + cos q )2 ( 4 sin2 q . cos2 q + 4 sinq .cos q + 1 − 1) 4 − cos 2 q d q (1 + sin 2q ) (1 + sin 2q )2 − 1

Now put 1 + sin 2q = t.

.

19

dq

p 2 p 2

20

61.

RUDIMENTS OF MATHEMATICS

I=

=

z

x 2 −1 dx x2

z

•let, x + 1 + a = z2

x+ +a x + +b 1 x

x

1 x

2 z dz z z + b −a 2

= 2

z

FG H

dz z + b− a

x2 + ax + 1 + x2 + bx + 1 +c x

= 2 log z + z2 + b − a + c = 2 log

62.

I=

=

z FH

dx x2 + 1

x+

IK 99 [let, x = tan q ; so, dx = sec

sec2 q d q

zb zb zb

secq + tan q

= 1 2

=

99

g

sec q

sec q + tan q

98

g

IJ K

then 2z dz = 1 − 12 dx— x

2

2

q dq ]

1 secq [(secq + tan q ) + (secq − tanq )]dq 99 2 secq + tanq

z

b

dq + 1 2

zb

g

secq d q secq + tan q

g100

sec q (sec q + tan q ) q )dq = 1 d q + 1 secq (secq + tan101 99 2 2 sec q + tan q secq + tanq

z

g

b

g

[since, sec q – tan q =

=–

LM MN

OP + c •where y = sec q + tan q = x + PQ

1 1 1 + 2 98 y98 100 y100

1 ] secq + tanq

x2 + 1 ,

and so, dy = sec q (sec q + tan q )d q — 63.

z tan

–1 (1 + 2x + 4x2 ) dx =

p x – 2

z tan p = x – z tan 2 =

1

−1

1+ 2 x + 4 x

2

z LNM

dx =

–1 (1 + 2x) dx +

OP Q

p − cot−1 (1 + 2 x + 4 x2 ) dx 2

px– 2

z tan

z tan

–1

1 + 2 x − 2 x dx 1 + 2 x (1 + 2 x )

–1 2x dx

[Now, let us put tan–1 2x = q . \

z tan

–1 2x dx =

z

q sec2 q dq = q tan q – 1 2 2 2

z tan q dq = q2 tan q – 12 log (sec q )

o SOLUTIONS OF * MARKED PROBLEMS (INTEGRAL CALCULUS)

= x tan–1 2x – log (1 + 4x2 )] p x 2x +1 = – tan–1 (2x + 1) + log {1 + (1 + 2x)2 } + x tan–1 2x – log(1 + 4x2 ). (Ans.) 2 2 p 4

z

65.

tann+1

q dq +

p 4

z

0

q dq =

tann–1

0

p 4

z

tann–1

0

p 2

Again, In = n q

z

sinn

q d q – (n – 1)

0 p 2

z

=–n

p 2

z q sin

p 4

L n O d q = M tan q P = 1 ... (1) N n Q0 n

n–2

q dq

0

q sinn–2 q (1 – sin2 q ) d q +

0

p 2

z q sin

n–2

q dq

0

p 2

z q cos q

=–n

q . sec2 q

(sinn–2

q . cos q ) d q +

0

p 2

z q sin

n–2

q dq

0 p 2

p 2

p 2

n−1 O n−1 L = – n Mq cosq sin q P + n z (cos q – q sin q ) sin q dq + z sinn–2 q d q n −1 Q n−1 N 0 0 0 L n O = n M sin q P − n z q sinn q d q + z q sinn–2 q d q n −1N n Q n−1 0 0 0 L O = 1 − 1 Mn z q sinn q d q − (n − 1) z q sinn−2 q dq P n−1 n−1M PQ 0 N0 1 I = 1 − 1 . In Þ FH1 + I = 1 Þ In = 1 ... (2) n−1 n−1 n − 1K n n − 1 n p 2

p 2

p 2

p 2

p 2

p 4

p 4

0

0

z

z

\ Thus, from (1) and (2), tann+1 q d q + tann–1 q dq p 2

=n q

z

sinn

q d q – (n – 1)

0

67.

z

=

z q sin

n–2

q dq .

0

dx = tan x + cot x + sec x + cosec x =

p 2

z z

=–

sin x cos x dx = 1 + sin x + cos x

z

z

sin x cos x

+

cos x sin x

dx +

1 cos x

sin x dx = sec x + tan x + 1

sin x (sec x − tan− 1) dx 1 1 =– dx + −2 tan x 2 2

z

z

+

1 sin x

sin x {sec x − (tan x + 1)}dx (sec x + tan x + 1)(sec x − tan x − 1)

z sinx dx + 12 z sec x dx

x 1 1 − cos x + log | sec x + tan x |. (Ans.) 2 2 2

21