1.

2m 1 Q estion2m – 1 Question4m – 1 QuestionTotal – 6 marks.

Definition: An equation whichcontains variables x & y,contains variables x & y,

Constants and derivatises like,

ll d d ff lis called  as a differentialequation.

Ex. 1) Ex. 1)

2)2)

ORDER OF A DIFFERENTIALEQUATION: It is the order of highestEQUATION: It is the order of highest order derivative present in the diff ti l tidifferential equation.

DEGREE OF A DIFFERENTIALDEGREE OF A DIFFERENTIALEQUATION: It is the highestpower of highest order derivativepresent in the differential equation.

Find the order and degree of theFind the order and degree of the following differential equations:

1))

Solution: Order = 2; degree = 3

2)

S l iSolution:

RP to 2 b.s

Order = 2; Degree =2.Degree   2.

F ti f diff ti l tiFormation of a differential equationby eliminating arbitrary constant(parameter)

1.

h iwhere c is a parameterSolution: Differentiating w.r.t.x 

2.

where a is a parameter

Solution: Differentiating w.r.t.x 

3. 

where c is a parameter

Solution: Differentiating w.r.t.x 

butbut

4. obtain the differential equation ofstraight lines passing through the origin.g

Soln : Eqn Of any line canSoln :  Eqn. Of any line canbe taken as  y = mx+cIf line passes through the originIf line passes through the origin,Then C=o

Therefore : eqn. of given family ofStraight lines can be taken as y = mx ;m is parameterConsider : y = mx …………………(1)Consider : y   mx …………………(1)

Diff w r t x =m(1)Diff  w.r.t x       =m(1)

Put m = in (1)

Y =  x

i.e.    x   ‐y = 0

5.5. Obtain the differential equation of Obtain the differential equation of family of circles touching yfamily of circles touching y axis ataxis atfamily of circles touching y family of circles touching y –– axis at axis at the origin.the origin.

ySolution :

X

Any circle in the family is passingthrough the  Origin therefore  c=0

Centre (‐g,‐f) of all the circlelies on x – axislies on x  axis

Therefore f = 0

Therefore equation of family of circleis : x2+y2+2gx=o: ……………… (1) 

Where  ‘g’  is parameter.g p

Diff w r t xDiff  w.r.t  x

2x+2y + 2g = 02x+2y + 2g = 0

Put 2g=‐2x‐ 2y In (1)Put 2g=‐2x‐ 2y In (1)

x2+y2+(‐2x‐2y )x = 0

x2+y2‐2x2‐2xy  = 0

Solution to first orderSolution to first order,first degree differential

ti b i blequations by variableseparable method.Express the given D.Eby separating the variablesy p gin the following form.

I t t bIntegrate b.s

Solve the following D. E

1

Integrate b.s 

2.

Solution:Solution:

Integrate b .s 

NNote: 

3.

Solution: 

= 0

M l i l b 2 bMultiply by 2 b.s

= 0

Integrate b.s 

c

Note:Note: 

44

Integrate b.s 

5.

Solution:-

Integrate with respect to the variablesIntegrate with respect to the variables

dx = dy

Log = log ( ) + C

6.6.

Given x=1; y=1, find the particular  solution.

Solution:-

(

--- 11

To find c

Put x=1, y=1 in ---1

-log1= c=-2 g

Put this in 1Particular solution is

77.+ = 0

Solution:-Solution:

Integrate b.s

88.

Solution:-Solution:

Integrate b.s

9.

Solution:

11

-y+1

-y-1

Integrate:

log

10. 10.

Solution:

integrateintegrate

1111.

Solution:

integrate

1212.

Solution:

integrate

put

Differential equations reducible tovariable separable form.Differential equations of the form.

Method : put

S lSolve 1)1)

Solution:

Put

Differentiating w. r. t. x

where

2. Solve  

Solution : put p

Differentiating w. r. t. x

integrating 

Area bounded by curvesArea bounded  by curvesT fi d b d d b f ( )To find area bounded by a curve y=f (x) on x-axis from x=a to x=b

Y = f (x)( )

X= a X=bX b

Problems :Find the area of the circle x2+y2=a2Find the area of the circle x +y =aby integration method.

Soln : x2+y2 = a2

y2 = a2 - x2y = a - x

By

B

0XA

X=a0 A

X=0X 0

T fi d th f t OABTo find the area of sector OAB This region is bounded by

2. Find   area of circle   x2+y2 = 8

Soln : Do the previous problem &in the last step put a2 = 8in the last step, put a = 8

3. To find the area of circle x2+y2 = 25

Soln : First find the area of circle x2+y2 = a2

& in the last step replace a2 by 25& in the last step replace a by 25

4.Find the area of Ellipse

by integration method.

yB

y

0 XA

X= a

X 0X=0

To find the area of sector OABThis region is bounded byg yThe curve

x‐axis  x=0 & x=a

5 Fi d th f th lli5. Find the area of the ellipseby integration.

Soln : First you do the previous problemi f llii.e. Area of ellipse = 

in this problem p

To find area boundedTo find area bounded between 2 curves

y Y2Y1

X = a X =bx

Problems :find the area bounded betweenthe following 2 curves.

1. Y2 = 4ax & x2 = 4aySolution : To find limitsSolution : To find limits

Y2 = 4ax……………… (1)2 4x2 = 4ay

X4 = 64a3 x4 3X4 ‐ 64a3 x=0

X (x3‐64a3)=0X=0        x3‐64a3 =0

x =4ax  4a

id Y2 4consider, Y2 = 4ax

Consider, x2 = 4ay

2. Y2 = x & x2 = y

Solution : To find limitsY2 = x……………… (1)Y  x……………… (1)x2 = yi e y = x2 use this in (1)i.e. y = x2 use this in (1)

x(x3‐1) =o  x =0, x  = 1

sq units

3. Y2 = 4x & y = xSolution : To find limits

Y2 = 4x……………… (1)y = xy = x in (1)

x =0  ; x  = 4

Properties of Definite IntegralsProperties of Definite IntegralsProperties of Definite IntegralsProperties of Definite Integrals

Part  Part  ‐‐ A   A   → 1 Mark → 1 Mark ‐‐1 Question1 QuestionPart  Part  ‐‐ D   D   → 6 Marks→ 6 Marks‐‐1 Question1 QuestionPart  Part  ‐‐ A   A   → 1 Mark → 1 Mark ‐‐1 Question1 QuestionPart  Part  ‐‐ D   D   → 6 Marks→ 6 Marks‐‐1 Question1 Question

Properties:Properties:11 ∫∫ bb f(x)f(x) dxdx == ∫∫ aa f (x)f (x) dxdx

Properties:Properties:11 ∫∫ bb f(x)f(x) dxdx == ∫∫ aa f (x)f (x) dxdx1.1.aa ∫ ∫ b b f(x) f(x) dxdx = = ‐‐ bb ∫ ∫ a a f (x) f (x) dxdxProof: Proof: Let Let aa ∫ ∫ bb f(x) f(x) dxdx = F(x)= F(x)1.1.aa ∫ ∫ b b f(x) f(x) dxdx = = ‐‐ bb ∫ ∫ a a f (x) f (x) dxdxProof: Proof: Let Let aa ∫ ∫ bb f(x) f(x) dxdx = F(x)= F(x)Consider, LHS = Consider, LHS = aa ∫ ∫ b b f(x) f(x) dxdx yy

= F(x) |= F(x) |aabb f(x)f(x)

Consider, LHS = Consider, LHS = aa ∫ ∫ b b f(x) f(x) dxdx yy

= F(x) |= F(x) |aabb f(x)f(x)aa f(x)f(x)

LHS = F(b)  LHS = F(b)  ‐‐ F(a)   …. 1F(a)   …. 1aa f(x)f(x)

LHS = F(b)  LHS = F(b)  ‐‐ F(a)   …. 1F(a)   …. 1

x = ax = a x = bx = bx = ax = a x = bx = b

Consider, RHS Consider, RHS  = = ‐‐ bb ∫ ∫ a a f (x) f (x) dxdxConsider, RHS Consider, RHS  = = ‐‐ bb ∫ ∫ a a f (x) f (x) dxdxbb

= = ‐‐ F(x) |F(x) |bbaa

== ‐‐ [F(a)[F(a) –– F(b)]F(b)]

bb

= = ‐‐ F(x) |F(x) |bbaa

== ‐‐ [F(a)[F(a) –– F(b)]F(b)]= =  [F(a) [F(a)  F(b)]F(b)]= = ‐‐ F(a) + F(b)F(a) + F(b)

RHSRHS F(b)F(b) F( ) 2F( ) 2

= =  [F(a) [F(a)  F(b)]F(b)]= = ‐‐ F(a) + F(b)F(a) + F(b)

RHSRHS F(b)F(b) F( ) 2F( ) 2RHS RHS  = F(b)  = F(b)  ‐‐ F(a)   …. 2F(a)   …. 2RHS RHS  = F(b)  = F(b)  ‐‐ F(a)   …. 2F(a)   …. 2

From 1 and 2; LHS = RHSFrom 1 and 2; LHS = RHSIeIe aa ∫ ∫ b b f (x) f (x) dxdx = = ‐‐ bb ∫ ∫ aa f (x) f (x) dxdx

From 1 and 2; LHS = RHSFrom 1 and 2; LHS = RHSIeIe aa ∫ ∫ b b f (x) f (x) dxdx = = ‐‐ bb ∫ ∫ aa f (x) f (x) dxdxaa ∫∫ ( )( ) bb ∫∫ ( )( )aa ∫∫ ( )( ) bb ∫∫ ( )( )

2. 2. aa ∫ ∫ b b f(x) f(x) dxdx = = aa ∫ ∫ c c f(x) f(x) dxdx + + cc ∫ ∫ b b f (x) f (x) dxdxWhere a < c < bWhere a < c < b2. 2. aa ∫ ∫ b b f(x) f(x) dxdx = = aa ∫ ∫ c c f(x) f(x) dxdx + + cc ∫ ∫ b b f (x) f (x) dxdxWhere a < c < bWhere a < c < bWhere a < c < bWhere a < c < bProof: Proof: LetLet ∫ ∫  f (x) f (x) dxdx = F(x)= F(x)ConsiderConsider LHSLHS ∫∫ bb f (x)f (x) dxdx

Where a < c < bWhere a < c < bProof: Proof: LetLet ∫ ∫  f (x) f (x) dxdx = F(x)= F(x)ConsiderConsider LHSLHS ∫∫ bb f (x)f (x) dxdxConsider, Consider,  LHS = LHS = aa ∫ ∫ b b f (x) f (x) dxdx

= F(x) |= F(x) |aabb yy

Consider, Consider,  LHS = LHS = aa ∫ ∫ b b f (x) f (x) dxdx= F(x) |= F(x) |aabb yy

LHS = F(b) LHS = F(b) –– F(a)  …1F(a)  …1 f(x)f(x)LHS = F(b) LHS = F(b) –– F(a)  …1F(a)  …1 f(x)f(x)

x = a      x = c   x = bx = a      x = c   x = bx = a      x = c   x = bx = a      x = c   x = b

RHS RHS  = = aa ∫ ∫ c c f (x) f (x) dxdx + + cc ∫ ∫ b b f (x) f (x) dxdx= F(x) |= F(x) |aac  c  + F(x) |+ F(x) |ccbb

RHS RHS  = = aa ∫ ∫ c c f (x) f (x) dxdx + + cc ∫ ∫ b b f (x) f (x) dxdx= F(x) |= F(x) |aac  c  + F(x) |+ F(x) |ccbb F(x) | F(x) |aa  F(x) | F(x) |cc= F(c) = F(c) –– F(a) + F(b) F(a) + F(b) –– F (c) F (c) 

RHSRHS = F(b)= F(b) F(a) 2F(a) 2

 F(x) | F(x) |aa  F(x) | F(x) |cc= F(c) = F(c) –– F(a) + F(b) F(a) + F(b) –– F (c) F (c) 

RHSRHS = F(b)= F(b) F(a) 2F(a) 2RHS RHS  = F(b) = F(b) –– F(a)    …2F(a)    …2From 1 and 2 ; LHS = RHSFrom 1 and 2 ; LHS = RHSRHS RHS  = F(b) = F(b) –– F(a)    …2F(a)    …2From 1 and 2 ; LHS = RHSFrom 1 and 2 ; LHS = RHS

ieieaa ∫ ∫ b b f(x) f(x) dxdx = = aa ∫ ∫ c c f(x) f(x) dxdx + + cc ∫ ∫ b b f (x) f (x) dxdxWhere a<c<bWhere a<c<bieieaa ∫ ∫ b b f(x) f(x) dxdx = = aa ∫ ∫ c c f(x) f(x) dxdx + + cc ∫ ∫ b b f (x) f (x) dxdxWhere a<c<bWhere a<c<b

3. 3. aa ∫ ∫ b b f (x) f (x) dxdx = = aa ∫ ∫ b b  f(t) f(t) dtdtff f( )f( ) ( )( )

3. 3. aa ∫ ∫ b b f (x) f (x) dxdx = = aa ∫ ∫ b b  f(t) f(t) dtdtff f( )f( ) ( )( )Proof: LetProof: Let ∫ f(x) ∫ f(x) dxdx = F(x)= F(x)

. . ... . ∫ f(t) ∫ f(t) dtdt = F(t)= F(t)

Proof: LetProof: Let ∫ f(x) ∫ f(x) dxdx = F(x)= F(x)

. . ... . ∫ f(t) ∫ f(t) dtdt = F(t)= F(t)

LHS LHS  = = aa ∫ ∫ b b f (x) f (x) dxdx = F(x) |= F(x) |aabb = F(b) = F(b) –– F(a) …1F(a) …1LHS LHS  = = aa ∫ ∫ b b f (x) f (x) dxdx = F(x) |= F(x) |aabb = F(b) = F(b) –– F(a) …1F(a) …1

RHS = RHS = aa ∫ ∫ b b f(t) f(t) dtdt = F(t) |= F(t) |aabb = F(b) = F(b) –– F(a) …2F(a) …2

From 1 and 2 ; LHS = RHSFrom 1 and 2 ; LHS = RHS

RHS = RHS = aa ∫ ∫ b b f(t) f(t) dtdt = F(t) |= F(t) |aabb = F(b) = F(b) –– F(a) …2F(a) …2

From 1 and 2 ; LHS = RHSFrom 1 and 2 ; LHS = RHSFrom 1 and 2 ; LHS = RHSFrom 1 and 2 ; LHS = RHSi.ei.e .  .  aa ∫ ∫ b b f (x) f (x) dxdx = = aa ∫ ∫ b b f (t) f (t) dtdt

From 1 and 2 ; LHS = RHSFrom 1 and 2 ; LHS = RHSi.ei.e .  .  aa ∫ ∫ b b f (x) f (x) dxdx = = aa ∫ ∫ b b f (t) f (t) dtdt

4. 4. 00∫∫a a f (x) f (x) dxdx = = 00∫∫aa f(af(a‐‐x) x) dxdx4. 4. 00∫∫a a f (x) f (x) dxdx = = 00∫∫aa f(af(a‐‐x) x) dxdx

= = aa∫∫oo f (af (a‐‐t) (t) (‐‐dtdt))== ‐‐ ∫∫oo f (af (a‐‐t)t) dtdt= = aa∫∫oo f (af (a‐‐t) (t) (‐‐dtdt))== ‐‐ ∫∫oo f (af (a‐‐t)t) dtdt= = ‐‐ aa∫∫ f (af (a‐‐t) t) dtdt= + = + oo∫∫aa f (af (a‐‐t) t) dtdt

∫∫aa f (f ( )) dd

= = ‐‐ aa∫∫ f (af (a‐‐t) t) dtdt= + = + oo∫∫aa f (af (a‐‐t) t) dtdt

∫∫aa f (f ( )) dd= = oo∫∫aa f (af (a‐‐x) x) dxdx=RHS=RHS= = oo∫∫aa f (af (a‐‐x) x) dxdx=RHS=RHS

Problems:Problems:Problems:Problems:

1.1. P.T   P.T    π/2    π/2     √√sinxsinx dxdx = = ππ√√cosxcosx + √+ √sinxsinx 44

1.1. P.T   P.T    π/2    π/2     √√sinxsinx dxdx = = ππ√√cosxcosx + √+ √sinxsinx 44

∫∫0  0   √√cosxcosx + √+ √sinxsinx 44

Solution: Solution: Let Let II = =  π/2    π/2     √√sinxsinx dxdx …1…1√√cosxcosx + √+ √sinxsinx

0  0   √√cosxcosx + √+ √sinxsinx 44Solution: Solution: Let Let II = =  π/2    π/2     √√sinxsinx dxdx …1…1

√√cosxcosx + √+ √sinxsinx∫∫

00 √√cosxcosx + √+ √sinxsinxChange x to π/2 Change x to π/2 ‐‐ x x II == π/2π/2 √sin(π/2√sin(π/2 x)x) dxdx

00 √√cosxcosx + √+ √sinxsinxChange x to π/2 Change x to π/2 ‐‐ x x II == π/2π/2 √sin(π/2√sin(π/2 x)x) dxdx

∫∫

∫∫II =     =     π/2    π/2     √sin(π/2 √sin(π/2 ––x)        x)         dxdx√√coscos(π/2 (π/2 ––x) + √sin(π/2 x) + √sin(π/2 ––x)x)

II =     =     π/2    π/2     √sin(π/2 √sin(π/2 ––x)        x)         dxdx√√coscos(π/2 (π/2 ––x) + √sin(π/2 x) + √sin(π/2 ––x)x)00∫∫

II /2/2II /2/2II = =  π/2    π/2     √√cosxcosx dxdx …2…2√√sinxsinx + √+ √cosxcosx

II = =  π/2    π/2     √√cosxcosx dxdx …2…2√√sinxsinx + √+ √cosxcosx00∫∫

Add 1 and 2Add 1 and 2II + + I=I= π/2  π/2   √√sinxsinx ++ √√cosxcosx dxdxAdd 1 and 2Add 1 and 2II + + I=I= π/2  π/2   √√sinxsinx ++ √√cosxcosx dxdx∫∫ √√cosxcosx + √+ √sinxsinx √√cosxcosx + √+ √sinxsinx2I=2I= π/2  π/2   √√sinxsinx ++ √√cosxcosx dxdx

√√cosxcosx + √+ √sinxsinx √√cosxcosx + √+ √sinxsinx2I=2I= π/2  π/2   √√sinxsinx ++ √√cosxcosx dxdx

00∫∫∫∫ √√cosxcosx + √+ √sinxsinx

2I=2I=00∫∫π/2 π/2 1 1 dxdx2I2I || /2/2 //

√√cosxcosx + √+ √sinxsinx2I=2I=00∫∫π/2 π/2 1 1 dxdx2I2I || /2/2 //

00∫∫2I=2I=x|x|ooππ/2/2 =  π/2 =  π/2 –– 00I = I = ππ/4/42I=2I=x|x|ooππ/2/2 =  π/2 =  π/2 –– 00I = I = ππ/4/4

2 ST2 ST π/2π/2 11 dxdx ππ2 ST2 ST π/2π/2 11 dxdx ππ∫∫2.ST 2.ST π/2 π/2  11 dxdx =  =  ππ

00 1+1+tantannnx 4x 4S l tiS l ti

2.ST 2.ST π/2 π/2  11 dxdx =  =  ππ

00 1+1+tantannnx 4x 4S l tiS l ti

∫∫Solution: Solution:

LetLet II π/2π/2 11 dd

Solution: Solution:

LetLet II π/2π/2 11 dd∫∫Let Let I I == π/2 π/2  1            1             dxdx00 1+ (1+ (sinsinnnxx//coscosnnxx))

Let Let I I == π/2 π/2  1            1             dxdx00 1+ (1+ (sinsinnnxx//coscosnnxx))∫∫

II == π/2 π/2  coscosnnxx dxdx ‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 11nn ++ ii nn

II == π/2 π/2  coscosnnxx dxdx ‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 11nn ++ ii nn

∫∫00 coscosnnxx + + sinsinnnxx

Change x to Change x to π π  ‐‐xx22

00 coscosnnxx + + sinsinnnxxChange x to Change x to π π  ‐‐xx

222222

coscosnn ππ –– x x  dxdxII == π/2 π/2  22

coscosnn ππ –– x x  dxdxII == π/2 π/2  22

00 ∫∫ coscosnn(π (π ––x)  + sinx)  + sinnn(π (π –– x)x)22 2 2 

coscosnn(π (π ––x)  + sinx)  + sinnn(π (π –– x)x)22 2 2 

0  0  ∫∫

II /2/2II /2/2∫∫II =  =   π/2 π/2  sinsinnnx          x           dxdx …2 …2 sinsinnnx + cosx + cosnnxx

II =  =   π/2 π/2  sinsinnnx          x           dxdx …2 …2 sinsinnnx + cosx + cosnnxx00∫∫

Take 1 + 2Take 1 + 2I +I = I +I = π/2 π/2  coscosnnx    x    ++ sinsinnnx             x             dxdxTake 1 + 2Take 1 + 2I +I = I +I = π/2 π/2  coscosnnx    x    ++ sinsinnnx             x             dxdx

00∫∫ coscosnnx+ sinx+ sinnnx    sinx    sinnnx + cosx + cosnnxx22I = I = π/2 π/2 coscosnnx+ sinx+ sinnnx   x    dxdx

coscosnnx+ sinx+ sinnnx    sinx    sinnnx + cosx + cosnnxx22I = I = π/2 π/2 coscosnnx+ sinx+ sinnnx   x    dxdx

00∫∫00∫∫ coscosnnx+ sinx+ sinnnx x 

22I = I = OO∫∫π/2 π/2 1dx1dx2I2I || /2/2 //

coscosnnx+ sinx+ sinnnx x 22I = I = OO∫∫π/2 π/2 1dx1dx2I2I || /2/2 //

00∫∫

2I=2I=x|x|ooπ/2π/2 =  π/2 =  π/2 –– 00I = I = π/4π/42I=2I=x|x|ooπ/2π/2 =  π/2 =  π/2 –– 00I = I = π/4π/4

3. ST 3. ST ∞∞ x          x           dxdx = = ππ(1+x) (1+x(1+x) (1+x22)) 44

3. ST 3. ST ∞∞ x          x           dxdx = = ππ(1+x) (1+x(1+x) (1+x22)) 4400∫∫ (1 x) (1 x(1 x) (1 x )) 44

Solution : Solution : LetLet I =I = ∞∞ xx dxdx

(1 x) (1 x(1 x) (1 x )) 44Solution : Solution : LetLet I =I = ∞∞ xx dxdx∫∫Let Let I I x          x           dxdx

(1+x) (1+x(1+x) (1+x22))Put x = Put x = tantanθθ ; θθ = tantan‐‐11xx

Let Let I I x          x           dxdx(1+x) (1+x(1+x) (1+x22))

Put x = Put x = tantanθθ ; θθ = tantan‐‐11xx

00∫∫θθ ; θθ

Diff Diff w.r.t.xw.r.t.x1 = sec1 = sec22θ  θ  dθdθ//dxdx

θθ ; θθDiff Diff w.r.t.xw.r.t.x1 = sec1 = sec22θ  θ  dθdθ//dxdx//dxdx = sec= sec22θ  θ  dθdθWhen x = 0; θ  = tanWhen x = 0; θ  = tan‐‐11 0 =00 =0

//dxdx = sec= sec22θ  θ  dθdθWhen x = 0; θ  = tanWhen x = 0; θ  = tan‐‐11 0 =00 =0;;When x = When x = ∞∞; θ  = tan; θ  = tan‐‐11 ∞∞ = π/2= π/2

;;When x = When x = ∞∞; θ  = tan; θ  = tan‐‐11 ∞∞ = π/2= π/2

II /2/2 θθ 22θθ dθdθII /2/2 θθ 22θθ dθdθI = I = π/2π/2 tantanθθ secsec22θ θ dθdθ(1+tan(1+tanθθ) (1+tan) (1+tan22θθ))/2/2 θθ 22θθ dθdθ

I = I = π/2π/2 tantanθθ secsec22θ θ dθdθ(1+tan(1+tanθθ) (1+tan) (1+tan22θθ))/2/2 θθ 22θθ dθdθ

00∫∫00∫∫= = π/2π/2 tantanθθ secsec22θ θ dθdθ

(1+tan(1+tanθθ) sec) sec22θθ/2/2 θθ dθdθ

= = π/2π/2 tantanθθ secsec22θ θ dθdθ(1+tan(1+tanθθ) sec) sec22θθ/2/2 θθ dθdθ

00∫∫= = π/2π/2 tantanθθ dθdθ

1+tan1+tanθθ/2/2 θθ// θθ dθdθ

= = π/2π/2 tantanθθ dθdθ1+tan1+tanθθ

/2/2 θθ// θθ dθdθ00∫∫

= = π/2π/2 sinsinθθ//cosθcosθ dθdθ1+ 1+ sinsinθθ//cosθcosθ

/2/2 θθ// θθ dθdθ

= = π/2π/2 sinsinθθ//cosθcosθ dθdθ1+ 1+ sinsinθθ//cosθcosθ

/2/2 θθ// θθ dθdθ00∫∫

= = π/2π/2 sinsinθθ//cosθcosθ dθdθ((coscosθ+sinθθ+sinθ)/)/cosθcosθ

= = π/2π/2 sinsinθθ//cosθcosθ dθdθ((coscosθ+sinθθ+sinθ)/)/cosθcosθ00∫∫

II /2/2 θθ dθdθ 11II /2/2 θθ dθdθ 11∫∫I = I = π/2π/2 sinsinθθ dθdθ …1…1coscosθθ+sin+sinθθ

θ /2θ /2 θθ

I = I = π/2π/2 sinsinθθ dθdθ …1…1coscosθθ+sin+sinθθ

θ /2θ /2 θθ00∫∫

Change Change θ to π/2 θ to π/2 -- θθI = I = π/2π/2 sin(sin(π/2 π/2 –– θ)θ) dθdθ

(( /2/2 θ)θ) (( /2/2 θ)θ)

Change Change θ to π/2 θ to π/2 -- θθI = I = π/2π/2 sin(sin(π/2 π/2 –– θ)θ) dθdθ

(( /2/2 θ)θ) (( /2/2 θ)θ)00∫∫ coscos((π/2 π/2 –– θ) θ) + sin(+ sin(π/2 π/2 –– θ)θ)I = I = π/2π/2 coscosθθ dθdθ …2…2

θθ θθ

coscos((π/2 π/2 –– θ) θ) + sin(+ sin(π/2 π/2 –– θ)θ)I = I = π/2π/2 coscosθθ dθdθ …2…2

θθ θθ

00∫∫00∫∫ sinsinθ+θ+coscosθθ

add 1 and 2add 1 and 2I II I /2/2 θθ θθ ddθθ

sinsinθ+θ+coscosθθadd 1 and 2add 1 and 2I II I /2/2 θθ θθ ddθθ

00∫∫

∫∫I + I = I + I = π/2   π/2    sinsinθθ ++ coscosθθ ddθθcoscosθθ+ + sinsinθθ sinsinθθ + + coscosθθ

I + I = I + I = π/2   π/2    sinsinθθ ++ coscosθθ ddθθcoscosθθ+ + sinsinθθ sinsinθθ + + coscosθθ00∫∫

2I2I /2/2 θθ θθ θθ2I2I /2/2 θθ θθ θθ2I = 2I = π/2 π/2  sinsinθθ +    +    coscosθθ ddθθsinsinθθ + + coscosθθ

2I2I /2/2 θθ

2I = 2I = π/2 π/2  sinsinθθ +    +    coscosθθ ddθθsinsinθθ + + coscosθθ

2I2I /2/2 θθ00∫∫

2I = 2I = OO∫∫ππ/2  /2  1d1dθθ2I=2I=θθ||ooππ/2/2 =  π/2 =  π/2 –– 00II /4/4

2I = 2I = OO∫∫ππ/2  /2  1d1dθθ2I=2I=θθ||ooππ/2/2 =  π/2 =  π/2 –– 00II /4/4I = I = ππ/4/4I = I = ππ/4/4

4 S T4 S T 114 S T4 S T 114. S.T 4. S.T 1 1  11 dxdx =    =    ππx + √1x + √1‐‐xx22 44

II 11

4. S.T 4. S.T 1 1  11 dxdx =    =    ππx + √1x + √1‐‐xx22 44

II 1100∫∫

Solution: Let Solution: Let I I =    =    1 1  1         1          dxdxx + √1x + √1‐‐xx22

θθ θ iθ i 11

Solution: Let Solution: Let I I =    =    1 1  1         1          dxdxx + √1x + √1‐‐xx22

θθ θ iθ i 1100∫∫

Put x= Put x= sinsinθθ ; θ = sin; θ = sin--11xxDiff Diff w.r.t.xw.r.t.x

θθ θθ//dd

Put x= Put x= sinsinθθ ; θ = sin; θ = sin--11xxDiff Diff w.r.t.xw.r.t.x

θθ θθ//dd1 = 1 = cosθcosθ. . ddθθ//dxdxdxdx = = coscosθθ dθdθhh θ iθ i 110 00 0

1 = 1 = cosθcosθ. . ddθθ//dxdxdxdx = = coscosθθ dθdθhh θ iθ i 110 00 0When x = 0; When x = 0; θ = sinθ = sin--110 = 00 = 0

When x = 1; When x = 1; θ = sinθ = sin--111 = π/21 = π/2When x = 0; When x = 0; θ = sinθ = sin--110 = 00 = 0When x = 1; When x = 1; θ = sinθ = sin--111 = π/21 = π/2

. . ... . I I =     =     π/2 π/2  11 coscosθθ ddθθ

sinsinθθ ++ √1√1 sinsin22θθ. . ... . I I =     =     π/2 π/2  11 coscosθθ ddθθ

sinsinθθ ++ √1√1 sinsin22θθ00∫∫00∫∫ sinsinθθ + + √1√1‐‐sinsin22θθ=    =    π/2  π/2  coscosθθ ddθθ

sinsinθθ + √cos+ √cos22θθ

sinsinθθ + + √1√1‐‐sinsin22θθ=    =    π/2  π/2  coscosθθ ddθθ

sinsinθθ + √cos+ √cos22θθ

00∫∫00∫∫00∫∫00∫∫ sinsinθθ + √cos+ √cos22θθ=    =    π/2  π/2  coscosθθ ddθθ …1…1

sinsinθθ ++ cosθcosθ

sinsinθθ + √cos+ √cos22θθ=    =    π/2  π/2  coscosθθ ddθθ …1…1

sinsinθθ ++ cosθcosθ

00∫∫00∫∫00∫∫00∫∫ sinsinθθ + + cosθcosθ

Change Change θ to π/2 θ to π/2 -- θ θ II == π/2π/2 coscos(π/2(π/2 θ)θ) ddθθ

sinsinθθ + + cosθcosθChange Change θ to π/2 θ to π/2 -- θ θ II == π/2π/2 coscos(π/2(π/2 θ)θ) ddθθ

00∫∫00∫∫

∫∫∫∫I I =    =    π/2  π/2   coscos(π/2 (π/2 –– θ)θ) ddθθsin(π/2 sin(π/2 –– θ) + θ) + coscos(π/2 (π/2 –– θ) θ)

I I =    =    π/2  π/2   coscos(π/2 (π/2 –– θ)θ) ddθθsin(π/2 sin(π/2 –– θ) + θ) + coscos(π/2 (π/2 –– θ) θ) 00∫∫00∫∫

II /2/2 θθ θθII /2/2 θθ θθ∫∫I I == π/2  π/2   sin sin θθ ddθθ …2…2

00 coscosθθ + + sinθsinθI I == π/2  π/2   sin sin θθ ddθθ …2…2

00 coscosθθ + + sinθsinθ∫ ∫ 

Add 1 and  2Add 1 and  2I+ I =I+ I = π/2 π/2  coscosθθ + + sinθsinθ dθdθ

θ θθ θ θθ i θi θ

Add 1 and  2Add 1 and  2I+ I =I+ I = π/2 π/2  coscosθθ + + sinθsinθ dθdθ

θ θθ θ θθ i θi θ∫ ∫ 

00 sinsinθ+cosθθ+cosθ coscosθθ ++sinθsinθ2I = 2I = π/2  π/2  coscosθθ ++sinθsinθ dθdθ

θ θθ θ

00 sinsinθ+cosθθ+cosθ coscosθθ ++sinθsinθ2I = 2I = π/2  π/2  coscosθθ ++sinθsinθ dθdθ

θ θθ θ

∫∫∫∫00 sinsinθ+cosθθ+cosθ

2I = 2I = 00∫∫π/2  π/2   dθdθ2I2I θθ|| /2/2 //

00 sinsinθ+cosθθ+cosθ2I = 2I = 00∫∫π/2  π/2   dθdθ2I2I θθ|| /2/2 //

∫ ∫ 2I=2I=θθ||ooπ/2π/2 =  π/2 =  π/2 –– 00I = I = ππ/4/42I=2I=θθ||ooπ/2π/2 =  π/2 =  π/2 –– 00I = I = ππ/4/4

Property 5Property 5‐‐

0∫2a f (x)dx =Property 5Property 5‐‐

0∫2a f (x)dx =20∫a f (x) dx ; If  f(2a‐ x)=f(x)

0 ; If f(2a‐ x) = ‐ f(x)

Proof: Consider LHS = 0∫2a f (x)dxf( ) 2 f( )

Proof: Consider LHS = 0∫2a f (x)dxf( ) 2 f( )

0      ; If  f(2a x)    f(x)

= 0∫a f(x)dx + a∫2a f(x) dxPut x= 2a –t ; t = 2a –x  in second integralff

= 0∫a f(x)dx + a∫2a f(x) dxPut x= 2a –t ; t = 2a –x  in second integralffDiff w.r.t.x

1 = ‐dt / dxd d h

Diff w.r.t.x1 = ‐dt / dxd d hdx = ‐dt When x= a; t=a

When x= 2a; t=odx = ‐dt When x= a; t=a

When x= 2a; t=o

= 0= 0∫∫aa f(x)f(x)dxdx + + aa∫∫00 f(2a f(2a ‐‐ t) (t) (‐‐dtdt))

= 0= 0∫∫aa f(x)f(x)dxdx –– aa∫∫00 f(2af(2a ‐‐ t)t) dtdt= 0= 0∫∫aa f(x)f(x)dxdx + + aa∫∫00 f(2a f(2a ‐‐ t) (t) (‐‐dtdt))

= 0= 0∫∫aa f(x)f(x)dxdx –– aa∫∫00 f(2af(2a ‐‐ t)t) dtdt= 0= 0∫∫ f(x)f(x)dxdx aa∫∫ f(2a f(2a  t) t) dtdt

= 0= 0∫∫aa f(x)f(x)dxdx + + 00∫∫aa f(2a f(2a ‐‐ t) t) dtdt

= 0= 0∫∫aa f(x)f(x)dxdx ++ 00∫∫aa f(2af(2a ‐‐ x)x) dxdx

= 0= 0∫∫ f(x)f(x)dxdx aa∫∫ f(2a f(2a  t) t) dtdt

= 0= 0∫∫aa f(x)f(x)dxdx + + 00∫∫aa f(2a f(2a ‐‐ t) t) dtdt

= 0= 0∫∫aa f(x)f(x)dxdx ++ 00∫∫aa f(2af(2a ‐‐ x)x) dxdx= 0= 0∫∫ f(x)f(x)dxdx   00∫∫ f(2a f(2a  x) x) dxdxCase Case ii: When f(2a : When f(2a ‐‐ x) = f(x)x) = f(x).. LHS =LHS = 00∫∫aa f(x)f(x)dxdx ++ 00∫∫aa f(x)f(x) dxdx

= 0= 0∫∫ f(x)f(x)dxdx   00∫∫ f(2a f(2a  x) x) dxdxCase Case ii: When f(2a : When f(2a ‐‐ x) = f(x)x) = f(x).. LHS =LHS = 00∫∫aa f(x)f(x)dxdx ++ 00∫∫aa f(x)f(x) dxdx. . .. LHS LHS  00∫∫ f(x)f(x)dxdx   00∫∫ f(x) f(x) dxdx

= 2= 200∫∫aa f(x) f(x) dxdx = RHS= RHSCase ii: When f(2a Case ii: When f(2a ‐‐ x) = x) = ‐‐ f(x)f(x)

. . .. LHS LHS  00∫∫ f(x)f(x)dxdx   00∫∫ f(x) f(x) dxdx= 2= 200∫∫aa f(x) f(x) dxdx = RHS= RHS

Case ii: When f(2a Case ii: When f(2a ‐‐ x) = x) = ‐‐ f(x)f(x)Case ii: When f( aCase ii: When f( a x)x) f(x)f(x)LHS =LHS = 00∫∫aa f(x)f(x)dxdx + + 00∫∫aa ‐‐ f(x) f(x) dxdx =  0  = RHS=  0  = RHSCase ii: When f( aCase ii: When f( a x)x) f(x)f(x)LHS =LHS = 00∫∫aa f(x)f(x)dxdx + + 00∫∫aa ‐‐ f(x) f(x) dxdx =  0  = RHS=  0  = RHS

∫∫∫∫5. P.T 5. P.T ππ x        x         dxdx = =  ππ22

00 1+sin1+sin22x         2√2x         2√25. P.T 5. P.T ππ x        x         dxdx = =  ππ

22

00 1+sin1+sin22x         2√2x         2√2Solution : Solution : 

II =  =  ππ x        x         dxdx …1…1Solution : Solution : 

II =  =  ππ x        x         dxdx …1…1∫∫00 1+sin1+sin22xx

Change x to πChange x to π‐‐xx00 1+sin1+sin22xx

Change x to πChange x to π‐‐xx

∫ ∫ 

∫∫=   =   ππ ππ ‐‐ x        x         dxdx

00 1+sin1+sin22(π (π –– x)x)=   =   ππ ππ ‐‐ x        x         dxdx

00 1+sin1+sin22(π (π –– x)x)

∫∫∫ ∫ =    =    ππ ππ ‐‐ x        x         dxdx …2…2

00 1+sin1+sin22 xx=    =    ππ ππ ‐‐ x        x         dxdx …2…2

00 1+sin1+sin22 xx∫ ∫ 

Add 1 and 2 Add 1 and 2 Add 1 and 2 Add 1 and 2 

2 I 2 I =       =       ππ x  + π x  + π ‐‐ x      x       dxdx1+sin1+sin22xx

2 I 2 I =       =       ππ x  + π x  + π ‐‐ x      x       dxdx1+sin1+sin22xx

∫ 00 1+sin1+sin22xx

=      =      ππ ππ dxdx1+sin1+sin22xx

00 1+sin1+sin22xx=      =      ππ ππ dxdx

1+sin1+sin22xx

∫∫

00 1+sin1+sin22xxDivide both Nr and Dr by cosDivide both Nr and Dr by cos22xx

= π= π ππ secsec22 xx dxdx

00 1+sin1+sin22xxDivide both Nr and Dr by cosDivide both Nr and Dr by cos22xx

= π= π ππ secsec22 xx dxdx

∫ 

∫= π    = π    ππ secsec22 x        x         dxdx

00 secsec22 x + tanx + tan22 xx= π= π ππ secsec22 xx dxdx

= π    = π    ππ secsec22 x        x         dxdx

00 secsec22 x + tanx + tan22 xx= π= π ππ secsec22 xx dxdx

∫ ∫= π     = π     ππ secsec22 x                 x                  dxdx

00 1+ 1+ tantan22 x+ tanx+ tan22 xx= π     = π     ππ secsec22 x                 x                  dxdx

00 1+ 1+ tantan22 x+ tanx+ tan22 xx∫ 

2 I2 I /2/2 222 I2 I /2/2 22∫2 I 2 I = π.= π.22 π/2π/2 secsec22 x           x            dxdx

00 1+ 21+ 2tantan22 xxII /2/2 22

2 I 2 I = π.= π.22 π/2π/2 secsec22 x           x            dxdx

00 1+ 21+ 2tantan22 xxII /2/2 22

∫ II=  π    =  π    ππ/2/2 secsec22 x           x            dxdx

00 1+(√2 1+(√2 tanxtanx))22II=  π    =  π    ππ/2/2 secsec22 x           x            dxdx

00 1+(√2 1+(√2 tanxtanx))22∫ Put √2 Put √2 tanxtanx = t= tDiff Diff w.r.t.xw.r.t.x

22 dd //dd

Put √2 Put √2 tanxtanx = t= tDiff Diff w.r.t.xw.r.t.x

22 dd //dd√2 sec√2 sec22 x     = x     = dtdt//dxdxsecsec22x x dxdx = 1/ √2 = 1/ √2 dtdthh

√2 sec√2 sec22 x     = x     = dtdt//dxdxsecsec22x x dxdx = 1/ √2 = 1/ √2 dtdthhWhen x=0 ; t=0When x=0 ; t=0

When x = π/2 ; t = When x = π/2 ; t = ∞∞When x=0 ; t=0When x=0 ; t=0When x = π/2 ; t = When x = π/2 ; t = ∞∞

II //II //∫∫I I =  =  ππ ∞∞ 1/√2 1/√2 dtdt

00 1+ 1+ tt22II

I I =  =  ππ ∞∞ 1/√2 1/√2 dtdt

00 1+ 1+ tt22II

∫ ∫ I I =  =  ππ ∞∞ 1 1 dtdt

√2√200 1+ 1+ tt22

11 ||

I I =  =  ππ ∞∞ 1 1 dtdt√2√200 1+ 1+ tt22

11 ||∫ ∫ 

= = ππ tantan‐‐11t |t |00∞∞√2√2

( /( / ))

= = ππ tantan‐‐11t |t |00∞∞√2√2

( /( / ))= =  ππ (π/2 (π/2 –– 0)0)√2√2

II

= =  ππ (π/2 (π/2 –– 0)0)√2√2

IIII =  =   ππII =  =   ππ

Property 6Property 6

aa∫∫aa f (x)f (x)dxdx == 2200∫∫a a f (x)f (x) dxdx ;; IfIf f(x) is evenf(x) is evenProperty 6Property 6

aa∫∫aa f (x)f (x)dxdx == 2200∫∫a a f (x)f (x) dxdx ;; IfIf f(x) is evenf(x) is even––aa∫∫ f (x)f (x)dxdx   2200∫∫ f (x) f (x) dxdx ; ; If If  f(x) is even  f(x) is even  0        ; 0        ; If If  f(x) is oddf(x) is odd

Proof: Consider LHS =Proof: Consider LHS = aa∫∫aa f (x)f (x)dxdx

––aa∫∫ f (x)f (x)dxdx   2200∫∫ f (x) f (x) dxdx ; ; If If  f(x) is even  f(x) is even  0        ; 0        ; If If  f(x) is oddf(x) is odd

Proof: Consider LHS =Proof: Consider LHS = aa∫∫aa f (x)f (x)dxdxProof: Consider LHS   Proof: Consider LHS   ‐‐aa∫∫ f (x)f (x)dxdx= = ‐‐aa∫∫0 0 f (x)f (x)dxdx + + 00∫∫aa f(x)f(x)dxdxPut x=Put x= ‐‐t in first integralt in first integral

Proof: Consider LHS   Proof: Consider LHS   ‐‐aa∫∫ f (x)f (x)dxdx= = ‐‐aa∫∫0 0 f (x)f (x)dxdx + + 00∫∫aa f(x)f(x)dxdxPut x=Put x= ‐‐t in first integralt in first integralPut x  Put x   t  in first integralt  in first integralDiff Diff w.r.t.xw.r.t.x1 = 1 = ‐‐dtdt / / dxdx

Put x  Put x   t  in first integralt  in first integralDiff Diff w.r.t.xw.r.t.x1 = 1 = ‐‐dtdt / / dxdxdtdt // dxdxdxdx = = ‐‐dtdt When x= When x= ‐‐a; t=aa; t=a

When x= 0; t=oWhen x= 0; t=o

dtdt // dxdxdxdx = = ‐‐dtdt When x= When x= ‐‐a; t=aa; t=a

When x= 0; t=oWhen x= 0; t=o;;;;

= a= a∫∫00 f(f(‐‐t) (t) (‐‐dtdt) + ) + 00∫∫aa f(x)f(x)dxdx

00∫∫aa f(f(‐‐t)t)dtdt ++ 00∫∫aa f(x)f(x)dxdx= a= a∫∫00 f(f(‐‐t) (t) (‐‐dtdt) + ) + 00∫∫aa f(x)f(x)dxdx

00∫∫aa f(f(‐‐t)t)dtdt ++ 00∫∫aa f(x)f(x)dxdx= 0= 0∫∫ f(f( t)t)dtdt + + 00∫∫ f(x)f(x)dxdx

= 0= 0∫∫aa f(f(‐‐x)x)dxdx + + 00∫∫aa f(x)f(x)dxdxCaseCase ii: Let f(x) be even function: Let f(x) be even function

= 0= 0∫∫ f(f( t)t)dtdt + + 00∫∫ f(x)f(x)dxdx

= 0= 0∫∫aa f(f(‐‐x)x)dxdx + + 00∫∫aa f(x)f(x)dxdxCaseCase ii: Let f(x) be even function: Let f(x) be even functionCase Case ii: Let f(x) be even function: Let f(x) be even function=> f(=> f(‐‐x) = f(x)x) = f(x).. LHS =LHS = 00∫∫aa f(x)f(x)dxdx ++ 00∫∫aa f(x)f(x) dxdx

Case Case ii: Let f(x) be even function: Let f(x) be even function=> f(=> f(‐‐x) = f(x)x) = f(x).. LHS =LHS = 00∫∫aa f(x)f(x)dxdx ++ 00∫∫aa f(x)f(x) dxdx. . .. LHS LHS  00∫∫ f(x)f(x)dxdx   00∫∫ f(x) f(x) dxdx

= 2= 200∫∫aa f(x) f(x) dxdx = RHS= RHSCase ii: Let f(x) be odd functionCase ii: Let f(x) be odd function

. . .. LHS LHS  00∫∫ f(x)f(x)dxdx   00∫∫ f(x) f(x) dxdx= 2= 200∫∫aa f(x) f(x) dxdx = RHS= RHS

Case ii: Let f(x) be odd functionCase ii: Let f(x) be odd functionCase ii: Let f(x) be odd functionCase ii: Let f(x) be odd function⇒⇒ f(f(‐‐x) = x) = ‐‐ f(x)f(x)⇒⇒ LHS =LHS = 00∫∫aa ‐‐f(x)f(x)dxdx ++ 00∫∫aa f(x)f(x) dxdx = 0 = RHS= 0 = RHS

Case ii: Let f(x) be odd functionCase ii: Let f(x) be odd function⇒⇒ f(f(‐‐x) = x) = ‐‐ f(x)f(x)⇒⇒ LHS =LHS = 00∫∫aa ‐‐f(x)f(x)dxdx ++ 00∫∫aa f(x)f(x) dxdx = 0 = RHS= 0 = RHS⇒⇒ LHS LHS  00∫∫ f(x)f(x)dxdx   00∫∫ f(x) f(x) dxdx   0    RHS    0    RHS  ⇒⇒ LHS LHS  00∫∫ f(x)f(x)dxdx   00∫∫ f(x) f(x) dxdx   0    RHS    0    RHS  

6. 6. ‐‐π/2π/2∫∫π/2π/2 sinsin33x x dxdx6. 6. ‐‐π/2π/2∫∫π/2π/2 sinsin33x x dxdx//

Solution: Solution: I I = = ‐‐π/2π/2∫∫π/2π/2 sinsin33x x dxdxLet f(x) = sinLet f(x) = sin33xx

//

Solution: Solution: I I = = ‐‐π/2π/2∫∫π/2π/2 sinsin33x x dxdxLet f(x) = sinLet f(x) = sin33xxPut x = Put x = ‐‐x x f(f(‐‐x) = x) = ‐‐ sinsin33xxPut x = Put x = ‐‐x x f(f(‐‐x) = x) = ‐‐ sinsin33xx

. . .... f(f(‐‐x) = x) = ‐‐f(x)f(x)

. . .... f(x) is odd function  => f(x) is odd function  => II = 0= 0

. . .... f(f(‐‐x) = x) = ‐‐f(x)f(x)

. . .... f(x) is odd function  => f(x) is odd function  => II = 0= 0