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Sample Math Homework | Sample Assignment | www.expertsmind.com

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

ò

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

ò coste dt t

Integrating by parts we get

ò udv = uv - ò vdu t

t

u= cost

V= e

du = -sint dt

dv= et dt V= et

u= sint du = Cost dt

ò e (-sint)dt cost + ò e (sint)dt

I= e cost I= et

dv= et dt

t

t

Further integrating by parts we get

I= et cost + et sint -

ò e (cost)dt t

I= et cost + et sint – I 2I= et cost + et sint 1 I= [ et cost + et sint] 2 1 I= [ eln(x) cos(ln(x)) + eln(x) sin(ln(x))] 2


1 [x cos(ln(x)) +xsin(ln(x))] 2 x I= [cos(ln(x))+sin(ln(x))]+C 2 I=

(b)

dy 1 = dx 3 + 5cos(x) dx 3 + 5cos(x) dx é 2 x ù ê 1 - tan 2 ú 3 + 5ê ú ê1 + tan2 x ú êë 2 úû é 2 xù ê1 + tan 2 údx ë û x x 3 + 3tan2 + 5 - 5tan2 2 2 2 x Sec dx 2 x 8 - 2tan2 2 x Sec 2 dx 2 xù é 2ê4 - tan2 ú 2 ë û

y=

ò

y=

ò

y=

ò

y=

ò

y=

ò

Put

tan

x = 2

t;

differentiating

1 x Sec 2 dx = dt 2 2 dt y= 4 - t2

ò

1 y= 4

(c)

ò

2-t +2+ t 1 é dt dt = ê + (2 - t)(2 + t) 4ë 2-t

dy = dx

ò

ò

dt ù 1 2 + t 1 = ln = ln 2 + t úû 4 2 - t 4

x 2 +c x 2 - tan 2

2 + tan

x 2 - 8x - 9

dy = x 2 - 8x + 16 - 16 - 9 dx dy = (x - 4)2 - 25 dx

y =

ò

(x - 4)2 - 25 dx

Put x-4= 5 coshu Differentiating dx= 5 sinhu du

y=

ò

ò

25cosh2u - 25 5sinhudu = 25sinh2udu

we

get


2

[

]

é eu - e -u ù 25 25 2u 25ê [e2u + e - 2u - 2]du = e - e - 2u - 4u ú du = 2 4 8 ëê ûú

y=

ò

y=

25 u e - e -u eu + e -u - 4u 8

y=

é 25 [4sinhucosh u - 4u] = 25 ê x - 4 8 2 ê 5 ë

ò

[(

)(

)

]

x 2 - 8x - 9 x - 4ù ú+c - cosh-1 5 5 ú û

Solution: (a)

ò ucoshudu

I=

dv= coshudu

t= u

V= sinhu

dt= du

ò tdv = tv - ò vdt = usinhu - ò sinhudu = usinhu - coshu + c dy y + = (x2 + 4)Cosh(x 2 + 4) dx x + 2 This is a linear differential equation. 1 P= and Q = (x2 + 4)Cosh(x 2 + 4) x+2 Integrating factor is

(b)

ò

dx

eò = e x + 2 = eln(x + 2) = (x + 2) Multiplying the given ODE by the integrating factor we get dy (x + 2) + y = (x + 2)(x2 + 4)Cosh(x 2 + 4) dx d(x + 2)y = (x + 2)(x 2 + 4)Cosh(x 2 + 4) dx Integrating on both sides pdx


ò (x + 2)(x

Y(x+2)=

ò x(x

2

+ 4)Cosh(x2 + 4)dx =

ò

2

+ 4)Cosh(x2 + 4)dx + 2 (x2 + 4)Cosh(x2 + 4)dx = I1+I2

In the first integral Put (x2+4) = t differentiating we get 2x dx= dt then it gets transformed to 1 1 1 I1= tcoshtdt = {tsinht -cosht}+C= {( x2+4)Sinh (x2+4)-cosh(x2+4)}+C 2 2 2 I2 is not a standard integral. This has to be done by series method. eu - e -u u2 u4 + ... , where * stands for factorial of a We know that Coshu= =1 + 2 2* 4* number. (x2 + 4)2 (x2 + 4)4 + ... In our case u= x2+4, replacing we get =Cosh (x2+4) 1 + 2* 4* (x2 + 4)3 (x2 + 4)5 + ... dx expanding and integrating we get I2= 1 + 2* 4*

ò

ò

I2=x+

1 x 7 12x 5 [ + + 24x 2 + 64x ]….. 2 7 5

(c)

Area under region p

p

-p

0

ò x coshx dx = 2ò x coshx dx

due to its symmetry about y- axis.

= 2[xsinhx - coshx ]0 = 2[psinhp-coshp +1] p

If the area density of the sheet is r, then the mass of the blade is 2r[psinhpcoshp +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 π

Then the center of mass=

òx

2

coshx dx

M

[x Sinhx - 2xcoshx - 2sinhx] =

p -p

2

M

(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)

d2 y dx

2

-4

dy +4=0 dx

equation: m2 - 4m + 4 = 0 ; (m-2)2=0; m=2 the roots are real and equal. Hence y=(Ax+B)e 2x Eigen vectors: Eigen values: d2 y dy (b) -4 + 4 = e x + 4x 2 dx dx Auxiliary

Auxiliary equation: m2 - 4m + 4 = 0 ; (m-2)2=0; m=2 the roots are real and equal. Hence General solution: y=(Ax +B)e 2x Particular Integral: 4x ex e x + 4x ex PI= = + = + 2 D - 4D + 4 D2 - 4D + 4 D 2 - 4D + 4 (1)2 - 4(1) + 4 4x 4(1 - D + D 2 / 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 Y= (2x +1)e 2x + ex + (x+1)

2x

+ ex + (x+1); we get

Solution: (a) I=

I=

ò [cos x] dx -1

2

Put x= Cost; differentiating we get dx= sint dt

ò t (sint)dt 2

dv= -sint dt

u= t2

V= cost

du=2tdt

dv= cost dt

u=2t

V= sint

du= 2dt

ò udv = uv - ò vdu I = t cost - ò 2tcostdt 2

I = t 2cost - I1 where I1=

ò 2tcostdt ò

I1 = 2tsint - 2sintdt I1 = 2tsint + 2cost I= t2cost-2tsint-2cost+c

[

]

2

I= x cos -1x -2 cos -1 x

{ x= Cost;

1 - x 2 -2x+C

(b) Trace of the given curve

1 - x 2 = sint ; t= cos -1 x }


Curve is symmetric about the Y axis p 2

Hence

òp

-

p 2

p

cosxdx = 2 cosxdx = 2[sinx ]02 =2[1-0]=2

ò 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. 1

1

Center of mass=

ò ρ2ycos

-1

ydy

0

M

=

ò ρ2ycos

-1

ydy

0

1

=

ò ycos

-1

ydy =. Integrating by

0

parts we get After substituting y = cost we get 1

ò 0

p 2

p 2

0

0

1 ycos ydy = cost sint tdt = 2 -1

ò

ò

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

Microsoft Equation 3.0

p

1 p tsin2tdt = [- 2tcos2t + sin2t]02 = 8 8

p ) 8


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