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Solution: (a) f(x)= Cos (ln(x))

I= ò dxCos(ln(x))

Put ln(x) = t X= et Differentiating dx = et dt Then

I= ò dtcostet

Integrating by parts we get

ò ò-= vduuvudv dv= et dt u= cost

V= et du = -sint dt

I= et cost - ò -sint)dt(et

I= et cost + ò (sint)dtet

Further integrating by parts we get dv= et dt u= sint V= et du = Cost dt

I= et cost + et sint - ò (cost)dtet

I= et cost + et sint – I 2I= et cost + et sint

I= 21

[ et cost + et sint]

I= 21

[ eln(x) cos(ln(x)) + eln(x) sin(ln(x))]

I= 21

[x cos(ln(x)) +xsin(ln(x))]

I=2x

[cos(ln(x))+sin(ln(x))]+C

(b) 5cos(x)3

1dxdy

+=

ò +=

5cos(x)3dx

y

ò

úúúú

û

ù

êêêê

ë

é

++

=

2x

tan1

2x

tan-153

dxy

2

2

ò-++

úû

ùêë

é +=

2x

5tan52x

3tan3

dx2x

tan1y

22

2

ò-

=

2x

2tan8

dx2x

Secy

2

2

òúû

ùêë

é -=

2x

tan42

dx2x

Secy

2

2

Put tan2x

= t; differentiating we get

dx2x

Sec21 2 = dt

ò -=

2t4

dty

dtt)t)(2(2t2t-2

41

y ò +-++

= = úû

ùêë

é+

+-ò ò t2

dtt2

dt41

=t2t2

ln41

-+

= c

2x

tan2

2x

tan2ln

41

+-

+

(c) 98xxdxdy 2 --=

916-168xxdxdy 2 -+-=

254)-(xdxdy 2 -=

ò -= dx 254)-(xy 2 Put x-4= 5 coshu

Differentiating dx= 5 sinhu du

5sinhudu 25u25coshy 2ò -= = ò udu25sinh2

du2ee

25y2uu

ò úúû

ù

êêë

é -=

-

= ò -+ - 2]due[e425 2u2u = [ ]4uee

825 2u2u -- -

( )( )[ ]4ueeee825

y uuuu -+-= --

[ ]4uu4sinhucosh825

y -= = c5

4xcosh

598xx

54x

225 1

2

+úú

û

ù

êê

ë

é --

--- -

Solution: (a)

I= òucoshudu dv= coshudu t= u

V= sinhu dt= du

ccoshuusinhu

sinhuduusinhu

vdttvtdv

+-=

-=

-=

òò ò

(b) 4)4)Cosh(x(x2x

ydxdy 22 ++=

++

This is a linear differential equation.

4)4)Cosh(x(x Q and 2x

1P 22 ++=

+=

Integrating factor is

2)(xeee 2)ln(x2xdx

pdx+=== ++òò

Multiplying the given ODE by the integrating factor we get

4)4)Cosh(x2)(x(xydxdy

2)(x 22 +++=++

4)4)Cosh(x2)(x(xdx

2)yd(x 22 +++=+

Integrating on both sides

Y(x+2)= ò +++ 4)dx4)Cosh(x2)(x(x 22 =

òò +++++ 4)dx4)Cosh(x(x24)dx4)Cosh(x(xx 2222 = I1+I2

In the first integral Put (x2+4) = t differentiating we get 2x dx= dt then it gets transformed to

I1= ò tcoshtdt21

= 21

{tsinht -cosht}+C= 21

{( x2+4)Sinh (x2+4)-cosh(x2+4)}+C

I2 is not a standard integral. This has to be done by series method.

We know that Coshu= 2ee uu --

= ...4u

2u

142

*+

*+ , where * stands for factorial of a

number.

In our case u= x2+4, replacing we get =Cosh (x2+4) ...4

4)(x2

4)(x1

4222

*+

+*+

+

I2= ò *+

+*+

+ dx ...4

4)(x2

4)(x1

5232

expanding and integrating we get

I2=x+21

[ 64x24x5

12x7x 2

57

+++ ]…..

(c)

Area under region

òò =pp

p 0-

dx coshx xdx coshx x 2 due to its symmetry about y- axis.

= [ ]p0coshxxsinhx2 - = 2[psinhp-coshp +1] If the area density of the sheet is r, then the mass of the blade is 2r[psinhp-coshp +1]. Center of mass Consider a thin strip of thickness dx at a distance of x from the origin as shown in the figure

Mass of the strip is rxcosh(x)dx

Then the center of mass= M

dx coshx xπ

π-

2ò=[ ]

M2sinhx2xcoshxSinhxx2 p

p--- (On

integrating by Parts successively) Center of mass = 0 as the numerator vanishes. The center of mass lies at the origin which is true from its symmetry.

Solution: (a) 04dxdy

4dx

yd2

2

=+-

Auxiliary equation: 044mm2 =+- ; (m-2)2=0; m=2 the roots are real and equal. Hence y=(Ax+B)e 2x

Eigen vectors: Eigen values:

(b) 4xe4dxdy

4dx

yd x2

2

+=+-

Auxiliary equation: 044mm2 =+- ; (m-2)2=0; m=2 the roots are real and equal. Hence General solution: y=(Ax +B)e 2x

Particular Integral:

PI= 44DD

4xe2

x

+-

+=

44DD

e2

x

+-+

44DD

4x2 +-

= 44(1)(1)

e2

x

+-+

)/DD-4(1

4x2 4+

=ex + (1-D+D2/4)-1(x)= ex + (x+1) Hence Y= GS+PI= y=(Ax +B)e 2x + ex + (x+1)

Y(0) = 3; hence 3=B+2; B= 1 Y’= (Ax+B) 2 e2x +A e2x +ex +1 At x= 0 Y’=6 6= 2B +A +2; A= 2 Substituting the values of A and B in Y= (Ax +B)e 2x + ex + (x+1); we get Y= (2x +1)e 2x + ex + (x+1)

Solution:

(a) I= [ ] dxxcos21ò - Put x= Cost; differentiating we get dx= sint dt

ò= (sint)dtt I 2 dv= -sint dt u= t2

V= cost du=2tdt

òò ò

-=

-=

2tcostdtcosttI

vduuvudv

2

12 I-costtI = where

I1= ò2tcostdt dv= cost dt u=2t

V= sint du= 2dt

2cost2tsintI

2sintdt2tsintI

1

1

+=

-= ò

I= t2cost-2tsint-2cost+c { x= Cost; sintx-1 2 = ; t= xcos 1- }

I= x [ ]21xcos- -2 xcos 1- 2x-1 -2x+C (b) Trace of the given curve

Curve is symmetric about the Y axis

Hence òò =

-

2

0

2

2

cosxdx2cosxdx

pp

p

= [ ]20sinx2p

=2[1-0]=2

(b) Consider a small strip of thickness dx at a distance of x from the origin as shown in the figure To find the position on the Y axis,

Consider a thin horizontal strip of thickness dy at a distance of y from the origin; the area of the strip is 2cos -1ydy. Hence the mass of the strip is r2cos -1ydy.

Center of mass=M

ydyρ2ycos 1-ò1

0 =2ρ

ydyρ2ycos1

0

1-ò= ò

1

0

1- ydyycos =. Integrating by

parts we get After substituting y = cost we get

ò1

0

1- ydyycos = ò2

0

tdt sint cost

p

= ò2

0

tsin2tdt21

p

= [ ]20sin2t2tcos2t81

p

+- =8p

Hence the center of mass of the strip is (0, 8p

)

Microsoft Equation 3.0