INTEGRATION OF TRIGONOMETRIC INTEGRALS

Recall the definitions of the trigonometric functions.

The following indefinite integrals involve all of these well-known

trigonometric functions. Some of the following trigonometry

identities may be needed.

A.)

B.)

C.) so that

D.) so that

E.)

F.) so that

G.) so that

It is assumed that you are familiar with the following rules of

differentiation.

These lead directly to the following indefinite integrals.

o 1.)

o 2.)

o 3.)

o 4.)

o 5.)

o 6.)

The next four indefinite integrals result from trig identities and u-

substitution.

o 7.)

o 8.)

o 9.)

o 10.)

We will assume knowledge of the following well-known, basic

indefinite integral formulas :

, where is a constant

, where is a constant

. Integrals with Inverse Trigonometric Functions

1.

2.

These formulas, if effectively, practiced would help you perform

well in mathematics section especially calculus part of your

boards and entrances, thus helping you secure good marks in

your class XII exams and helping you secure a good rank in

entrance exams.

How to score well

Before you start the exam, utilize the first 15 minutes to scan

the paper. Read the question paper thoroughly before

jumping to write the answers.

Among the questions with internal choices, select the ones

that you plan to attempt, and frame skeletons of the answers

you are going to write for these questions.

Follow a pattern. For example, in case you start with long

answer questions, complete that section and only then move

to short or very short answer section.

Highlight the important points and write your answer in

points to enhance visibility.

Points to remember

Marks are deducted for

missing steps. So

remember to write down

all the steps.

Practice. Practice. Practice.

This is the mantra for

scoring good marks in CBSE

Class 12 Mathematics

Exam.

Make NCERT book your

bible. Revise and practise

all the problems solved in

the NCERT book.

Question: Integrate . Let

u = x-1

so that

du = (1) dx = dx .

In addition, we can “back substitute” with

x = u+1 .

Substitute into the original problem, replacing all forms of x,

getting

Question: Integrate . Let

u = 2x+3

so that

du = 2 dx ,

or

(1/2) du = dx .

In addition, we can “back substitute” with

x = (1/2)(u-3) .

Substitute into the original problem, replacing all forms of x,

getting

Question: Integrate . Let

u = x+2

so that

du = (1) dx = dx .

In addition, we can “back substitute” with

x = u-2 .

Substitute into the original problem, replacing all forms of x,

getting

Question: Integrate . Let

so that

.

In addition, we can “back substitute” with

.

Substitute into the original problem, replacing all forms of x,

getting

Question: Integrate . Use u-substitution. Let

u = 1+3e-x

so that (Don’t forget to use the chain rule on e-x.)

du = 3e-x(-1) dx = -3e-x dx ,

or

(-1/3)du = e-x dx .

However, how can we replace the term e-3x in the original problem

? Note that

.

From the u-substitution

u = 1+3e-x ,

we can “back substitute” with

e-x = (1/3)(u-1) .

Substitute into the original problem, replacing all forms of x,

getting

(Recall that (AB)C = AC BC .)

Question: Integrate . Use u-substitution. Let

u = e2x+6ex+ 1

so that (Don’t forget to use the chain rule on e2x.)

du = (2e2x+6ex) dx

= (2ex+x+6ex) dx

= (2exex+6ex) dx

= 2ex(ex+3) dx

= 2ex(3+ex) dx

or

(1/2) du = ex(3+ex) dx .

Substitute into the original problem, replacing all forms of x,

getting

(Do not make the following very common mistake :

. Why is this incorrect ?)

.

Question: Integrate . First, factor out e9x

from inside the parantheses. Then

(Recall that (AB)C = AC BC .)

(Recall that (AB)C = ABC .)

.

Now use u-substitution. Let

u = 27+e3x

so that (Don’t forget to use the chain rule on e3x.)

du = 3e3x dx ,

or

(1/3) du = e3x dx .

Substitute into the original problem, replacing all forms of x , and

getting

.

Question: Integrate . Use u-substitution. Let

so that

,

or

.

Substitute into the original problem, replacing all forms of ,

getting

.

Question: Integrate . First multiply by

, getting

.

.

.

Now use u-substitution. Let

so that

.

Substitute into the original problem, replacing all forms of ,

getting

.

Question: Integrate . Let

and

so that

and .

Therefore,

.

SOLUTION 6 : Integrate . Let

and

so that (Don’t forget to use the chain rule when differentiating

.)

and .

Therefore,

.

Now use u-substitution. Let

so that

,

or

.

Then

+C

+C

+C.

Question: Integrate . Let

and

so that

and .

Therefore,

(Add in the numerator. This will replicate the

denominator and allow us to split the function into two parts.)

.

Question: Integrate . Let

and

so that

and .

Therefore,

.

Integrate by parts again. Let

and

so that

and .

Hence,

.

SOLUTION : Integrate . Use the power

substitution

so that

,

,

and

.

Substitute into the original problem, replacing all forms of ,

getting

.

SOLUTION 6 : Integrate . Use the power

substitution

so that

,

,

and

.

Substitute into the original problem, replacing all forms of ,

getting

(Use polynomial division.)

.

Question: Integrate . Because the degree of the

numerator is not less than the degree of the denominator, we

must first do polynomial division. Then factor and decompose

into partial fractions, getting

(After getting a common denominator, adding fractions, and

equating numerators, it follows that ;

let ;

let .)

(Recall that .)

Question: Integrate . Because the degree of

the numerator is not less than the degree of the denominator,

we must first do polynomial division. Then factor and

decompose into partial fractions, getting

(After getting a common denominator, adding fractions, and

equating numerators, it follows that ;

let ;

let .)

.

SOLUTION : Integrate . Use the power substitution

so that

and

.

Substitute into the original problem, replacing all forms of ,

getting

(Use polynomial division.)

.

SOLUTION 4 : Integrate . Use the power

substitution

so that

,

,

and

.

Substitute into the original problem, replacing all forms of ,

getting

.

SOLUTION : Integrate . Use the power substitution

so that

and

.

Substitute into the original problem, replacing all forms of ,

getting

(Use polynomial division.)

.

Use the method of partial fractions. Factor and decompose into

partial fractions, getting

(After getting a common denominator, adding fractions, and

equating numerators, it follows that ;

let ;

let .)

(Recall that .)

.

Question: Integrate . Decompose into partial fractions,

getting

(After getting a common denominator, adding fractions, and

equating numerators, it follows that ;

let ;

let

;

it follows that and .)

.

Question: Integrate . Use u-substitution. Let

so that

.

Now rewrite this rational function using rules of exponents. Then

.

Substitute into the original problem, replacing all forms of ,

getting

.

Question: Integrate . First complete the square

in the denominator, getting

.

Now use u-substitution. Let

so that

.

In addition, we can “back substitute” with

.

Substitute into the original problem, replacing all forms of ,

getting

.

In the first integral use substitution. Let

so that

,

or

.

Substitute into the first integral, replacing all forms of , and use

formula 3 from the beginning of this section on the second

integral, getting

Integrate . First, use polynomial division to divide

by . The result is

.

In the second integral, use u-substitution. Let

so that

.

Substitute into the original problem, replacing all forms of ,

getting

(Now use formula 1 from the introduction to this section.)

.

SOLUTION : Integrate . Let

and

so that

and .

Therefore,

.

Use integration by parts again. let

and

so that

and .

Hence,

.

To both sides of this “equation” add , getting

.

Thus,

(Combine constant with since is an arbitrary

constant.)

.

Question: Integrate . Use integration by

parts. Let

and

so that

and .

Therefore,

.

Use integration by parts again. let

and

so that

and .

Hence,

.

From both sides of this “equation” subtract ,

getting

.

Thus,

(Combine constant with since is an arbitrary constant.)

INTEGRATION AS LIMIT OF A SUM AND ITS EVALUATION

Question: Use the limit definition of definite integral to evaluate

Divide the interval into equal parts each of length

for . Choose the sampling points to be the right-

hand endpoints of the subintervals and given by

for . The function is

.

Then the definite integral is

(Use summation rule 6 from the beginning of this section.)

(Use summation rules 5 and 1 from the beginning of this section.)

(Use summation rule 2 from the beginning of this section.)

.

Application of Integrals

Q. 1. Find the area of the region in the first quadrant

enclosed by the x-axis, the line y = x and the circle

Q. 2. Find the area of the region bounded by the ellipse

.

Q. 3. Find the area of the region bounded by the parabola

y = x2 and y = .

Q. 4. Find the area of the smaller part of the circle x2 + y2 =

a2 cut off by the linex= .

Q. 5. Using integration, find the area of the region bounded

by the triangle whose vertices are (1, 0), (2,2) and (3, 1).

Q. 6. Prove that the curves y2 = 4x and x2 = 4y divide the

area of the square bounded by x=0, x=4, y=4 and y=0 into

three equal parts.

Q. 7. Sketch the graph of y=

Q. 8. Using the method of integration, find the area

bounded by the curve .

Q. 9. Find the area of the smaller region bounded by the

ellipse .

Q. 10. Using integration, find the area of the triangular

region, the equations of whose sides are y=2x + 1, y=3x +1

and x = 4.

Q. 11. Find the area of the region

Q. 12. Find the area of the region between the circles x2 +

y2 = 4 and (x – 2)2 + y2 = 4.

Q. 13. Find the area bounded by the ellipse and

2 2 2

the co-ordinates x = ae and x = 0, where b =a (1 – e ) and

e<1.

Q. 14. Find the area bounded by the curve y2 = 4a2(x – 1)

and the lines x = 1and y = 4a.

Q. 15. Using integration, find the area of the region

bounded by the following curves, after making a rough

sketch: y = 1 +

Q. 16. Draw a rough sketch of the curves y = sinx and y =

cosx as x varies from o to and find the area of the

region enclosed by them and x-axis.

Q. 17. Find the area lying above x-axis and included

between the circle x2+ y2 = 8x and the parabola y2 = 4x.

Q. 18. Using integration find the area of the triangular

region whose sides have the equations y = 2x + 1, y = 3x +

1 and x = 4.

Q. 19. Find the area enclosed between the parabola y2 =

4ax and the line y = mx.

Q. 20. Find the area of the region bounded by the

parabolas y2 = 4 ax and x2 = 4 by

(NCERT)

Question 12: [ use: sin2x = 1- cos2x, ans. is x – sinx+c]

Question 14: [ Use: 1+sin2x= (cosx+sinx)2 , put

cosx+sinx = t , ans.is -1/(cosx+sinx)+c ]

Question 18: dx [use: cos2x = cos²x-sin²x , ans.tanx+c]

Question 22: [multiply & divide by sin(a-b), write

sin(a-b) = sin{(x-b)-(x-a)} in Nr., use formula of sin(A-B).ans. is

+c ]

Question 23: is equal to [(A) ]

A. tan x + cot x + C B. tan x + cosec x + C C. − tan x + cot x + C

D. tan x + sec x + C

Question 24: equals [(B) ]

A. − cot (exx) + C B. tan (xex) + C C. tan (ex) + C D. cot (ex) + C

Question 5: * Put x² = t , ans. is (3/2√2)tan-1(√2X²) +C +

Question 9: [Put tanx = t, ans. is log|tanx+ |+c]

Question 14: [ Dr. Can be written as =

= , ans. is sin-1( )+c ]

Question 17: [ + dx, put x²-1=t in 1st

integral, ans. is +2 log|x+ | +c ]

Question 18: [ let 5x-2 = P.d/dx(1+2x+3x²)+Q, P=5/6 &

Q=-11/3 , Ans. is 5/6 log (1+2x+3x²) – (11/3√2) tan-1( ) +c]

Question 25: equals *Dr. √–(4x²-9x)⇨ )² ,

(B)]

A B C D

Question 3: [ by partial fraction, A/(x-1)+B/(x-

2)+C/(x-3) ⇨A=1,B=-5 & C=4, ans. is log|x-1|-5log|x-2+4log|x-3|+c]

Question 8: [ A/(x-1)+B/(x-1)2 +C/(x+2)⇨ A=-C=2/9,

B=1/3, Ans. is 2/9log| |-1/3(x-1) +c]

Question 10: [same as Q.3, A=-1/10,B=5/2 & C=-24/5,

Ans. is 5/2 log|x+1|-1/10log|x-1|-12/5log|2x+3|+c ]

Question 12: [after division, we get x+ ⇨A=1/2,

B=3/2 , ans. is x²/2+1/2log|x+1|+3/2log|x-1|+c ]

Question 15: [ A/(x+1)+B/(x-1)+Cx+D/(x²+1)⇨ A=-

1/4,B=1/4,C=0 & D=-1/2 , ans. is ¼ log| |-1/2 tan-1x+c ]

Question 17: [Hint: Put sin x = t, ans.is log| |+c]

Question 18: [ put x²=y , , after dividing , we

get , 1- , by partial fraction A/(y+3) +B/(y+4) ⇨ A=-1, B=3,

ans. is x+(2/√3)tan-1(x/√3)-3tan-1(x/2) +c ]

Question 23: A. [(A) , multiply &

2

divide by x, put x = t, by partial fraction.]

B. C. D.

Question 5: x log 2x [integral by parts, (log2x).x²/2- ) dx

⇨ (log2x).x²/2 – x²/4+c ]

Question 14: [ integral by parts, (logx)².x²/2-

2

.1/x](x /2)dx , again by parts ⇨ (logx)².x²/2- [log x.(x²/2)-

] ⇨ (logx)².x²/2- x2/2(logx)+(1/4)x2 +c]

Question 6: [ ⇨ –

(9/2)log|(x+2)+ +c ]

Question 7: [ )2 ] ⇨(2x-3)/4

+(13/8)sin-1 (2x-3)/√3 +c +

Question 20: [ , by parts

⇨xex -ex – 4/п[cos(пx/4)] at x=0 to 1⇨ 1+4/п – 2√2/п ]

Question 4: [ ans. is 16/15(2+√2) +

Question 6: [ Dr. (17/4)- (x-1/2)² ⇨ (1/√17)log| |

Put x=0 to 2 ⇨ (1/√17) log( )]

Question 8: [by parts , ans is (e2/4)(e²-2) ]

Question 9:The value of the integral is A. 6 B. 0 C. 3 D. 4

-1

[ put x=sinѲ , limit will change from Ѳ=sin 1/3 when x=1/3 &

Ѳ=п/2 when x=1, ⇨ dѲ , put cotѲ=t, again

limit will change from √8 to 0 ans. is (A) =6]

IMPORTANT PROPERTIES OF DEFINITE INTEGRALS:

1. =

2. = a<c<b

3. =

4. =

5. = +

6. =2 , if f(2a-x) = f(x)

= 0 , if f(2a-x) = – f(x)

7. = 2 , if f is an even function i.e., f-x) = f(x)

= 0, if f is an odd function i.e., f(-x) = -f(x).

Question 6: [ ⇨ 9]

Question 10: [

=

⇨ –

=- ]

Question 12: [use property 4.⇨2I = . =2п ]

Question 15: [ use property 4. ⇨ 2I = =0 ]

Question 16: [ use property 4. ⇨ 2I =

⇨ I= =2 (BY PROP.6)= 2I1 …..(i) , where I1

= (by using prop. 4) ⇨ 2I1 =

⇨ – = I2 – п/2.log2 …..(ii) , where I2

= = (put 2x=y) = (by

prop. 6), from (i) & (ii) we get I = – п log2. ]

Question 19: Show that if f and g are defined

as and

[ by prop. 4 ⇨ I = = =

(according to given part) ⇨ I = 2 ]

Question 21:The value of is A. 2 B. C. 0 D.

[ use prop. 4 ⇨ 2I = =0 ]

Misc. Question 3: [Hint:Put ,ans. Is -2/a +c]

Question 4: [ put x= , ans. Is – ]

Question 5:

* ans. Is 2√x- 3×1/3 +6×1/6 –

6log|x1/6 +1| +c ]

Question 10: [ Nr. can be written as

2 2

(1-2sin xcos x)(-cos2x), ans. Is -1/2 sin2x +c ]

Question 11: [ same as Ques. 22]

Question 15: [ put cosx =y , ans. Is -1/4 cos4x +c ]

Question 16: [ put t = x4+1 , ans. Is ¼ log| x4+1| +c ]

Question 18: [ Dr. = sin4x cos + sin3xcosxsin =

sin4x(cos + cotxsin ) ( by using formula of sin(A+B)) , put t= cos

+ cotxsin , ans. Is -2/sin . +c ]

Question 19: [ use sin-1√x + cos-1√x =п/2 ⇨

2/ dx – x , put √x= t, integrate by parts & use

formula of dt , ans. Is 4/п{ sin-1√x+ } –x +c]

Question 20: [ put x2 = cosy , use cos2y= (1+cos2y)/2, ans. Is

-2 √(1-x) + cos-1√x + √x . √(1-x) +c ]

Question 21: [use (f(x)+f’(x))dx , sin2x = 2sinx.cosx &

2 x

1+cos2x = 2cos x , ans. Is e tanx +c ]

Question 22: [ by partial fraction , we get A/(x+2) +

B/(x+1) +C/(x+1)2 ⇨ A=3, B=-2 & C=1, ans. Is log – (X+1)-1 +c ]

*Question 24: [ after simplification,we

integrate . log(1+ ) dx , put x = tanѲ , then put sinѲ=t (by

parts) , ans. Is -1/3 (1+ )3/2 { log(1+ )-2/3 } +c ]

Question 25: [ same as Ques. 21, ans. Is eп/2 ]

Question 26: [ divide Nr. & Dr. by cos4x , put tan2x =

y & limit will change from 0 to 1, ans. Is п/8 ]

Question 27: [ use sin2x = 1 – cos2x , Nr. Can be

written as 4-3cos2x-4 ⇨ -п/6+ 4/3 dx , put tanx = t,

limit will change from 0 to ∞, ans. Is п/6 or we can do it by another

method ( by partial fraction) divide Nr. & Dr. by cos2x , put tanx = t]

Question 28: [ put sinx-cosx=t ∵ sin2x=1-(sinx-cosx)2

Limit will change from –(√3-1)/2 to (√3-1)/2 ⇨ ∵ even

fn. , ans. Is 2 ]

Question 30: [ put sinx-cosx = t, same as Ques. 28 ,

limit will change from -1 to 0, ans. Is 1/40 log9 ]

Question 31: [ use sin2x formula , put sinx=t ,

integrate by parts , ans. Is п/2 -1 ]

Question 32: [ use prop. 4 ⇨ 2I = dx=

– , ans. Is ]

Question 33: [ +

+ dx+ dx + dx = 19/2 ]

Question 34: [ by partial fraction A/x +B/x2

+C/(x+1) ⇨ A= -1, B=1 & C=1 ]

Question 39: [ by parts ∫1. Sin-1xdx ]

Question 40: Evaluate dx as a limit of sum.

[nh =1, = +………+f((n-1)h)]

+………….. ]

. ( ∵ nh=1) =( ( =1 ]

Question 43:If then is equal to

A. B. C. D.

Question 44:The value of is A. 1 B. 0 C. – 1 D.

[Nr.=x+(x-1) & Dr.=1-x(1-x), use prop. 4 , it gives tan-1x+tan-1(x-1),B]

Definite Integral

Question 2: Find the area of the region bounded by y2 = 9x, x = 2, x

= 4 and the x-axis in the first quadrant.

Question 5:Find the area of the region bounded by the ellipse

Question 6:Find the area of the region in the first quadrant

enclosed by x-axis, line and the circle

Question 7:Find the area of the smaller part of the circle x2 + y2 = a2

cut off by the line

Question 9:Find the area of the region bounded by the parabola y =

x2 and

Question 10:Find the area bounded by the curve x2 = 4y and the line

x = 4y – 2

Question 1:Find the area of the circle 4×2 + 4y2 = 9 which is interior

to the parabola x2 = 4y

Question 2:Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 +

y2 = 1

Question 5:Using integration find the area of the triangular region

whose sides have the equations y = 2x +1, y = 3x + 1 and x = 4.

Question 4:Using integration finds the area of the region bounded

by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).

Question 6:Smaller area enclosed by the circle x2 + y2 = 4 and the

line x + y = 2 is

A. 2 (π – 2) B. π – 2 C. 2π – 1 D. 2 (π + 2)

Question 7:Area lying between the curve y2 = 4x and y = 2x is

A. B. C. D.

Question 4:Sketch the graph of and evaluate

Question 8:Find the area of the smaller region bounded by the

ellipse and the line

Question 11:Using the method of integration find the area bounded

by the curve

[Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x

+ y = 1 and – x – y = 11]

Question 12:Find the area bounded by curves

Question 14:Using the method of integration find the area of the

region bounded by lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0

Question 15:Find the area of the region

Question 17:The area bounded by the curve , x-axis and the

ordinates x = –1 and x = 1 is given by [Hint: y = x if x > 0 and y = –x2

2

if x < 0]

A. 0 B. C. D.

Question 18:The area of the circle x2 + y2 = 16 exterior to the

parabola y2 = 6x is

A. B. C. D.

Question 19:The area bounded by the y-axis, y = cos x and y = sin x

when

Integration Problems

1. ∫(2×3 + 5x + 1)e2x dx [ans. e2x(x3 – 3/2×2 + 4x – 3/2) + C ]

2. ∫cos2 x tan2 x dx [ x/2− 1/4 sin 2x + C]

3. ∫e cos 2x

sin x cos x dx [ (−1/4) ecos 2x + C]

4. dx [ ln |2 + tan x| + C]

5. [ x2/2 − 3x + 8 ln |x + 3| + C]

6. dx [ 1/3(x2 + 4)3/2 − 4(x2 + 4)1/2 + C]

7. dx [ 2/3 ln |1 + 3√x| + C]

8. dx [x +ln|x|+1/2ln|x2+4|−1/2arctan(x/2)+c]

9. [2/3 sin3x – cosx+c]

10. [ (2 – √2)/3 ]

**Question: Integrate . Let

.

In addition, we can “back substitute” with

,

or

x = (4-u)2 = u2-8u+16 .

Then

dx = (2u-8) du .

In addition, the range of x-values is

,

so that the range of u-values is

,

or

.

Substitute into the original problem, replacing all forms of x,

getting

.

INTEGRATION AS LIMIT OF A SUM AND ITS EVALUATION

Question: Divide the interval into equal parts each of

length

for . Choose the sampling points to be the right-

hand endpoints of the subintervals and given by

for . The function is

.

Then the definite integral is

(Use summation rule 6 from the beginning of this section.)

(Use summation rules 5 and 1 from the beginning of this section.)

(Use summation rule 2 from the beginning of this section.)

.

.

SOLUTION : Divide the interval into equal parts each of

length

for . Choose the sampling points to be the right-

hand endpoints of the subintervals and given by

for . The function is

.

Then the definite integral is

(Use summation rule 6 from the beginning of this section.)

(Use summation rules 5 and 1 from the beginning of this section.)

(Use summation rule 2 from the beginning of this section.)

.

Question: Divide the interval into equal parts each of length

for . Choose the sampling points to be the right-

hand endpoints of the subintervals and given by

for . The function is

.

(Use L’Hopital’s rule since the limit is in the indeterminate form of

.)

.

**Question: Integrate . First, split the function into

two parts, getting

(Recall that .)

(Use formula 2 from the introduction to this section on integrating

exponential functions.)

.

**SOLUTION : Integrate . First, use polynomial

division to divide by . The result is

.

In the third integral, use u-substitution. Let

so that

,

or

.

For the second integral, use formula 2 from the introduction to

this section. In the third integral substitute into the original

problem, replacing all forms of , getting

(Now use formula 1 from the introduction to this section.)

**Question: Integrate . First rewrite this

rational function as

.

Now use u-substitution. Let

.

so that

,

or

.

In addition, we can “back substitute” with

.

Substitute into the original problem, replacing all forms of ,

getting

=

**Question: Integrate . Because the degree of the

numerator is not less than the degree of the denominator, we

must first do polynomial division. Then factor and decompose

into partial fractions (There is a repeated linear factor !), getting

(After getting a common denominator, adding fractions, and

equating numerators, it follows that

;

let ;

let ;

let

;

let

;

it follows that and .)

**Question: Integrate . Factor and

decompose into partial fractions (There is a repeated linear factor

!), getting

(After getting a common denominator, adding fractions, and

equating numerators, it follows that

;

let ;

let ;

let ;

let

.)

.

**SOLUTION : Integrate . Factor and decompose

into partial fractions (There are two repeated linear factors !),

getting

(After getting a common denominator, adding fractions, and

equating numerators, it follows that

;

let ;

let ;

let

;

let

;

it follows that and .)

.

**Question: Integrate . Begin by rewriting the

denominator by adding , getting

(The factors in the denominator are irreducible quadratic factors

since they have no real roots.)

(After getting a common denominator, adding fractions, and

equating numerators, it follows that

;

let

;

it follows that and and ;

let

it follows that and and .)

.

Now use the method of substitution. In the first integral, let

so that

.

In the second integral, let

so that

.

In addition, we can “back substitute”, using

in the first integral and

in the second integral. Now substitute into the original problems,

replacing all forms of , getting

(Recall that .)

.

**Solution: Integrate. U se the power substitution

so that

and

.

Substitute into the original problem, replacing all forms of ,

getting

(Use polynomial division.)

.

**Question: Integrate . Because we want to

simultaneously eliminate a square root and a cube root, use the

power substitution

so that

,

,

,

and

.

Substitute into the original problem, replacing all forms of ,

getting

(Use polynomial division. PLEASE INSERT A FACTOR OF 6 WHICH

WAS ACCIDENTLY LEFT OUT.)

.

**SOLUTION : Integrate . Remove the “outside”

square root first. Use the power substitution

so that

,

,

,

and (Use the chain rule.)

.

Substitute into the original problem, replacing all forms of ,

getting

.

**SOLUTION : Integrate . Remove the cube root

first. Use the power substitution

so that

,

,

,

and (Use the chain rule.)

.

Substitute into the original problem, replacing all forms of ,

getting

.

**Question: Integrate . Remove the

“outside” square root first. Use the power substitution

so that

,

and (Use the chain rule.)

,

or

.

Substitute into the original problem, replacing all forms of ,

getting

,

and

.

Substitute into the original problem, replacing all forms of ,

getting

.

Use the method of partial fractions. Factor and decompose into

partial fractions, getting (There are repeated linear factors!)

(After getting a common denominator, adding fractions, and

equating numerators, it follows that

;

let ;

let ;

let

;

let

;

it follows that and .)

(Recall that .)

.

**SOLUTION : Integrate . Use the power

substitution

so that

and

.

Substitute into the original problem, replacing all forms of ,

getting

(Use polynomial division.)

.

Use the method of partial fractions. Factor and decompose into

partial fractions, getting

(After getting a common denominator, adding fractions, and

equating numerators, it follows that

;

let ;

let ;

let

;

it follows that and and .)

.

DEFINITE INTEGRAL

Theory:

To find the area between two intersecting curves that only

intersect at two points, we first find the ‘x’ coordinates of the two

intersection points: x = a and x = b. Definite integrals give us the

area under each curve from x = a to b, then we subtract the two

areas to obtain the area between the curves. In the diagram below,

the area between the two graphs is shaded:

Area under a Curve

The area between the graph of y = f(x) and the x-axis is given by the

definite integral below. This formula gives a positive result for a

graph above the x-axis, and a negative result for a graph below the

x-axis.

Note: If the graph of y = f(x) is partly above and partly below the x-

axis, the formula given below generates the net area. That is, the

area above the axis minus the area below the axis.

Formula:

Example Find the area between y = 7 – x2 and the x-

1: axis between the values x = –1 and x = 2.

Example Find the net area between y = sin x and the

2: x-axis between the values x = 0 and x = 2π.

Area between Curves

The area between curves is given by the formulas below.

Formula 1:

for a region bounded above and below by y =

f(x) and y = g(x), and on the left and right by x =

a and x = b.

Formula 2:

for a region bounded left and right by x = f(y)

and x = g(y), and above and below by y = c and y

= d.

Example 1:1 Find the area between y = x and y = x2 from x = 1

to x = 2.

Example 2:1 Find the area between x = y + 3 and x = y2 from y

= –1 to y = 1.

Area Under a Curve

Definite Integrals

So far when integrating, there has always been a constant term

left. For this reason, such integrals are known as indefinite

integrals. With definite integrals, we integrate a function between

2 points, and so we can find the precise value of the integral and

there is no need for any unknown constant terms [the constant

cancels out].

The Area Under a Curve

The area under a curve between two points can be found by doing

a definite integral between the two points.

To find the area under the curve y = f(x) between x = a and x = b,

integrate y = f(x) between the limits of a and b.

Areas under the x-axis will come out negative and areas above the

x-axis will be positive. This means that you have to be careful when

finding an area which is partly above and partly below the x-axis.

You may also be asked to find the area between the curve and the

y-axis. To do this, integrate with respect to y.

Example

Find the area bounded by the lines y = 0, y = 1 and y = x2.

EXAMPLE 4: Find the area between the curve f (x) = cos п x

on the interval [0, 2].

SOLUTION:

STEP 1: Graph the function.

(See figure 3)

STEP 2: Set up the integrals

and evaluate.

Notice that the area we

have to find is in three figure 3

pieces. The intervals [0, .5]

and [1.5, 2] are above the x-

axis, and the interval [.5,

1.5] is below. Therefore, we

will need to have three

integrals. Also notice that

symmetry cannot be used in

this problem.

EXAMPLE 5: Find the area between the curves f (x) = 4 – x 2

and g (x) = x 2 – 4.

SOLUTION:

STEP 1: Graph the functions.

(See figure 4)

The reason for graphing the

two equations is to be able

to determine which function

is on top and which one is

on the bottom. Sometimes, figure 4

you can also determine the

points of intersection. From

this graph, it is cleat that f

(x) is the upper function, g

(x) is the lower function, and

that the points of

intersection are x = -2 and x

= 2.

STEP 2: Determine the points of intersection.

If you did not determine the points of intersection from the graph,

solve for them algebraically or with your calculator. To find them

algebraically, set each equation equal to each other.

4 – x 2 = x 2 – 4 ⇨ -2x 2 = -8 ⇨ x 2 = 4 ⇨ x = -2 or x = 2

STEP 3: Set up and evaluate the integral.

Recall from early in the notes, when we were finding the area

between the curve and the x-axis, we had to determine the upper

and the lower curve. Then the area was defined to be the following

integral.

So the definite integral would be the following.

Now, let us evaluate the integral.

If you look at the graph of the two functions carefully, you should

have noticed that we could have used some symmetry when

setting up the integral. The region is symmetric with respect to

both the x- and the y-axis. If we had used the y-axis symmetry, the

resulting integral would have had bounds of 0 and 2, and we would

have had to take 2 times the area to find the total area. Here is this

integral.

If we had used both symmetries, the resulting integral would still

have bounds of 0 and 2, but the upper function would have been f

(x) and the lower function would be y = 0 (the x-axis). To find the

total area, we would have to take this area times 4. Here is this

integral.

EXAMPLE 7: Find the area between the curves x = y 3 and x = y 2

that is contained in the first quadrant.

SOLUTION: STEP 1: Graph the functions. (See figure 6)

Since both equations are x in terms of y, we will integrate with

respect to y. When integrate with respect to x, we have to

determine the upper function and the lower function. Now that we

are integrating with respect to y, we must determine what function

is the farthest from the y-axis. The function that is the farthest

from the y-axis is x = y 2. So that will be our upper curve. The lower

curve will be the curve that is nearest to the y-axis. In this case, it is

the function x = y 3.

figure 6

STEP 2: Find the points of intersection.

Set the two equations equal to each other.

y 2 = y 3 ⇨ y 2 – y 3 = 0 ⇨ y 2 (1 – y) = 0 ⇨ y = 0 or y = 1

STEP 3: Set up and evaluate the integral.

using definite integrals to find the area between two curves

From the figure we can easily get that the area of the shaded

portion spqr = area tpqu – area tsru.

This is equivalent to the area enclosed between the curve y = f(x),

the x-axis and the ordinates x=a and x = b Minus the area

enclosed between the curve y = g(x), the x-axis and the ordinates

x = a and x = b. this is expressed mathematically as follows:

a

Therefore, the area between the two curves can be expressed as

a

Example – 3

Find the area bounded by the curves y = x2 and y = 2x.

Solution:

Step 1: To find the region we need to sketch the graph and find

where the two curves intersect.

To find where the curves intersect, we will set them equal to

each other and solve for x.

2x = x2

X2 – 2x = 0

X(x – 2) = 0

X = 0 or x – 2 = 0

X = 0 or x = 2

Plugging x = 0 into y = 2x gives us y = 2(0) = 0

Plugging x = 2 into y = 2x gives us y = 2(2) = 4

Therefore, the curves intersect at the points (0, 0) and (2, 4)

Step 2: as we can see in the figure, we are to find the area of the

shaded portion oabdo.

Area oabdo = area of oabco – area of odbco.

= the area enclosed between the straight line y = 2x, x-axis, x = 0

and

X = 2 Minus the area enclosed between the curve y = x2, x-axis, x

= 0 and x = 2.

Step 3: solve the definite integral.

square units

Example – 4:

Find the area bounded by the curves x2 = 4y and y2 = 4x.

Solution:

Step 1: Solve the given equations to find the points of

intersection.

(1) x2 = 4y, (2) y2 = 4x

Squaring both sides of (1) gives us x4 = 16y2

Substituting y2 = 4x into this equation gives us

x4 = 16(4x)

x4 = 64x

x4 – 64x = 0

x(x3 – 64) = 0

x = 0 or x3 = 64

x = 0 or x = 4

Plugging x = 0 into x2 = 4y gives us 0 = 4y implies that y = 0

Plugging x = 4 into x2 = 4y gives us 16 = 4y implies that y = 4

therefore, the points of intersection are (0, 0) and (4, 4)

Step 2: Sketch the graph.

Step 3: Solve both equations for y and write the formula for

finding the area of the shaded region.

Y2 = 4x Y = 2

since this is the equation of the top line, this will be the first part

of our equation.

X2 = 4y Y = x2

since this is the equation of the bottom line, this will be the

second part of our equation.

(recall the formula )

Therefore, the area of the shaded portion Sq. Units

Area Bounded by Two Curves: See Figure 12.3-8.

Example 1

Find the area of the region bounded by the graphs of f (x)=x3 and

g(x )=x. (See Figure 12.3-9.)

Step 1. Sketch the graphs of f (x ) and g (x ).

Step 2. Find the points of intersection.

Step 3. Set up integrals.

Note: You can use the symmetry of the graphs and let area

Analternate solution is to find the area using a calculator. Enter

and obtain .

Example 2

Find the area of the region bounded by the curve y =ex, the y-axis

and the line y =e2.

Step 1. Sketch a graph. See Figure 12.3-10.

Step 2. Find the point of intersection. Set e2 =ex x =2.

Step 3. Set up an integral:

Or using a calculator, enter and obtain (e2 +1).

Example 3

Using a calculator, find the area of the region bounded by y = sin x

and between 0≤ x ≤ π.

Step 1. Sketch a graph. See Figure 12.3-11.

Step 2. Find the points of intersection.

Using the [Intersection] function of the calculator, the

intersection points are x =0 and x =1.89549.

Step 3. Enter nInt(sin(x ) &8211; .5x, x, 0, 1.89549) and obtain

0.420798 ≈ 0.421.

(Note: You could also use the function on your calculator

and get the same result.)

Example 4

Find the area of the region bounded by the curve x y =1 and the

lines y = –5, x =e, and x =e3.

Step 1. Sketch a graph. See Figure 12.3-12.

ASSIGNMENT OF INTEGRATION

Question 1 Evaluate: (i)** Integrate .[ Use the power

substitution

Put ]

** (iii) Integrate . [ Use the power substitution

Put ]

(iii) [answer is (2 – √2)/3 ]

(iv) ∫ dx[multiply÷ by sin(a-b)](v) dx

[multiply & divide by ] (Vi)∫ dx [by partial fraction]

(v) dx [ use ∫ex(f(x)+f’(x))dx+ (vi) dx [put sinx=

, cosx = , then put t=tanx/2. Answer is – ]

(vii) dx [ + = ∫+ve dx+∫ -ve dx ,

answer is 5/2п- 1/п2] (viii) [ write sin2x = 1-cos2x answer is

п/6] (ix) + dx * answer is √2 ] (x) dx [ put

x=atan2Ѳ , answer is a/2(п-2) ] (xi) dx [ use property dx

= dx , dx = dx ∵f(2a-x) = f(x) , then put t=tanx,

answer is п²/2√2 ] (xii) dx , where f(x) =|x|+|x+2|+|x+5|.

[ dx + dx , answer is 31.5 ] (xiii) Evaluate dx

[use (f(x)+f’(x))dx Question 2 Using integration, find the area of the

regions: (i) { (x,y): |x-1| ≤y ≤ }

(ii) *(x,y):0≤y≤x2+3; 0≤y≤2x+3; 0≤x≤3+

[(i) A= dx- dx – dx = 5/2 [

+ ] – ½ ] [(ii) dx + dx , answer is 50/3]

(iii) Find the area bounded by the curve x 2 = 4y & the line x = 4y – 2.

[A = dx – dx = 9/8 sq. Unit.]

**(iv) Sketch the graph of f(x) = ,evaluate dx

[hint: dx = dx + dx = 62/3.]

**Question 3 evaluate dx [ mult. & divide by , put 1+x

=A.(d/dx)(x2+x)+B ,find A=B=1/2, integrate]

Definite integral as the limit of a sum , use formula : dx

, where nh=b-a & n→∞ Question 4 Evaluate

) dx (ii) dx

[ use = 1 for part (i) , use formulas of special sequences, answer

is 6]

Some special case :

(1) Evaluate: [ put x+1=t²] (2) [ put x+1 = t² ]

(3) Evaluate: (4) Evaluate: [ put x=1/t for both]

(5) Evaluate: [ divide Nr. & Dr. By x2 , then write x²+1/x²=(x-1/x)²

+2 according to Nr. , let x-1/x=t]

(6) Evaluate dx [ let x=A(d/dx) ( 1+x-x²) +B]

(7) Integrating by parts evaluate =

(8) Evaluate dx = dx [ put

sinx=Ad/dx(sinx+cosx)+B(sinx+cosx)+C

If Nr. Is constant term then use formulas of sinx,cosx as Ques. No. 1 (vi) part]