integrals with inverse trigonometric functions

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

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.

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

o 10.)

We will assume knowledge of the following well-known, basic indefinite integral formulas :

, where is a constant

. Integrals with Inverse Trigonometric Functions

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

u = 2x+3

so that

du = 2 dx ,

(1/2) du = dx .

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

u = x+2

so that

du = (1) dx = dx .

x = u-2 .

so that

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 ,

(-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) .

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

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

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

(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 ,

(1/3) du = e3x dx .

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

so that

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

getting

Question: Integrate . First multiply by

, getting

. Now use u-substitution. Let

so that

getting

and so that

and . Therefore,

SOLUTION 6 : Integrate . Let

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

and . Therefore,

so that

. Then

and so that

and . Therefore,

(Add in the numerator. This will replicate the denominator and allow us to split the function into two parts.)

so that

and . Therefore,

. Integrate by parts again. Let

and so that

and . Hence,

SOLUTION : Integrate . Use the power substitution

so that

getting

SOLUTION 6 : Integrate . Use the power substitution

so that

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 .)

(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

let .)

so that

getting

SOLUTION 4 : Integrate . Use the power substitution

so that

getting

so that

getting

. Use the method of partial fractions. Factor and decompose into

partial fractions, getting

let .)

(Recall that .)

Question: Integrate . Decompose into partial fractions, getting

it follows that and .)

so that

. Now rewrite this rational function using rules of exponents. Then

getting

Question: Integrate . First complete the square in the denominator, getting

so that

. In addition, we can "back substitute" with

getting

. In the first integral use substitution. Let

so that

. 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

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.)

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 the co-ordinates x = ae and x = 0, where b2=a2(1 – e2) 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: ∫ cos2 x+2sin ² xcos ² x 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

1sin(a−b)

log∨cos ( x−a )cos ( x−b )

∨¿ +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+√ tan ² x+4

Question 14: [ Dr. Can be written as√8−(x ²−3 x )

√8−(x2−2. 32 x+ 92 )+ 94 = √ 414 −(x−32 )² , ans. is sin-1(2x−3√ 41

Question 17: [ ∫ x

√ x ²−1dx

+ ∫ 2

√ x ²−1 dx, put x²-1=t in 1st integral, ans. is √ x ²−1 +2 log|x+√ x ²−1| +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(3x+1√2 ) +c]

Question 25: equals [Dr. √–(4x²-9x)⇨√ 8116−¿¿) , (B)]²

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| x−1x+2|-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+2 x+1

( x+1 )( x−1) ⇨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| x−1x+1|-1/2 tan-1x+c ]

Question 17: [Hint: Put sin x = t, ans.is log|2−sinx1−sinx|+c]

Question 18: [ put x²=y , ( y+1 )( y+2)( y+3 )( y+4) , after dividing , we

get , 1-2(2 y+5)

( y+3 )( y+4) , 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 & divide by x, put x2 = t, by partial fraction.]

B. C. D.

Question 5: x log 2x [integral by parts, (log2x).x²/2-∫ 12 x.2¿) dx⇨ (log2x).x²/2 – x²/4+c ]

Question 14: [ integral by parts, (logx)².x²/2-∫¿¿.1/x](x2/2)dx , again by parts ⇨ (logx)².x²/2- [log x.(x²/2)-∫ 1x . x ² /2dx ] ⇨ (logx)².x²/2- x2/2(logx)+(1/4)x2 +c]

Question 6: [ √ ( x+2 )²−(3 ) ² ⇨ x+22

√x ²+4 x−5 –(9/2)log|(x+2)+√ x ²+4 x−5 +c ]

Question 7: [ √¿¿)2 ] ⇨(2x-3)/4√1+3 x−x ²+(13/8)sin-1 (2x-3)/√3 +c ]

Question 20: [ ∫01

x ex dx+∫0

sinπx4dx

, 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|√172

+(x−12 )√172

−(x−12)|

Put x=0 to 2 ⇨ (1/√17) log(21−5√174 ) ]

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 [ put x=sinѲ , limit will change from Ѳ=sin-11/3 when x=1/3 &

Ѳ=п/2 when x=1, ⇨ ∫sin−1 1

cot5 /3θ .cos²θ d , put cot =t, again Ѳ Ѳlimit will change from √8 to 0 ans. is (A) =6]IMPORTANT PROPERTIES OF DEFINITE INTEGRALS:

1. ∫a

f ( x )dx = −∫b

f ( x )dx

2. ∫a

f ( x )dx = ∫a

f ( x )dx+∫c

f (x )dx a<c<b

3. ∫a

f ( x )dx = ∫a

f (a+b−x )dx

4. ∫0

f ( x )dx = ∫0

f (a− x )dx

5. ∫0

f ( x )dx = ∫0

f ( x )dx +∫0

f (a− x )dx

6. ∫0

f ( x )dx = 2∫0

f ( x )dx , if f(2a-x) = f(x)

= 0 , if f(2a-x) = - f(x)

7. ∫−a

f (x )dx = 2∫0

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

∫−a

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

Question 6: [∫2

−(x−5)dx+∫5

(x−5)dx ⇨ 9]

Question 10: [

[2 logsinx¿−log 2−logsinx−logcosx]dx ¿ =

logsinxdx−( log2)∫0

1dx−∫0

logcos( π2−x )dx ⇨ ∫

logsinxdx− (log 2 )π /2-

logsinxdx =-π2log 2]

Question 12: [use property 4.⇨2I = ∫0π

dx1+sinx .

1−sinx1−sinx =2п ]

Question 15: [ use property 4. ⇨ 2I =∫0π /2

0 =0 ]

Question 16: [ use property 4. ⇨ 2I = ∫0

log (1−cos ² x )dx

⇨ I = ∫0

logsinxdx =2∫0

logsinxdx (BY PROP.6)= 2I1 …..(i) , where I1 =∫0

logsinxdx

(by using prop. 4) ⇨ 2I1 = ∫0

log (2 sinx . cosx ) /2dx

⇨ ∫0

logsin2 xdx - ∫0

log2dx = I2 - п/2.log2 …..(ii) , where I2 =∫0

logsin2 xdx =

1/2∫0

logsinydy (put 2x=y) = ∫0

logsinydy (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 = ∫0

f ( x )g ( x )dx = ∫0

f (a− x ) g (a−x )dx = ∫0

f ( x )[4−g ( x )]dx

(according to given part) ⇨ I = 2 ∫0

f ( x )dx ]

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

[ use prop. 4 ⇨ 2I = ∫0

log1dx =0 ]

Misc. Question 3: [Hint:Put ,ans. Is -2/a√ ax−1 +c]

Question 4: [ put x= √ tant

, ans. Is – (1+x4)

Question 5:

[ ans. Is 2√x- 3x1/3 +6x1/6 -6log|x1/6 +1| +c ]

Question 10: [ Nr. can be written as (1-2sin2xcos2x)(-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 cosα

+ sin3xcosxsinα

= sin4x(cosα + cotxsinα ) ( by using formula of sin(A+B)) , put t= cosα +

cotxsinα , ans. Is -2/sinα .√ sin (x+α)sinx +c ]

Question 19: [ use sin-1√x + cos-1√x =п/2 ⇨ 2/π ∫2sin−1√ x dx – x , put √x= t, integrate by parts & use formula of

∫√1−t ² dt , ans. Is 4/п{ 2x−12 sin-1√x+ √ x √ (1−x )2 } –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 ∫ ex

(f(x)+f’(x))dx , sin2x = 2sinx.cosx & 1+cos2x = 2cos2x , ans. Is extanx +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+2)3

(x+1)2- (X+1)-1 +c ]

*Question 24: [ after simplification,we

integrate √x ²+1x4

. log(1+ 1x ² ) dx , put x = tanѲ , then put sinѲ=t (by

parts) , ans. Is -1/3 (1+ 1x ² )

3/2 { log(1+ 1x ² )-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 ⇨ - /6+ 4/3п ∫0

π /2sec ² x

4 sec ² x−3dx , 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 ⇨ 2 ∫0

√3−12

dt√1−t ²

∵ even fn.

, ans. Is 2 sin−1 √3−12 ]

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 = π∫0π

tanxsecx+tanx dx=

π ² - π∫0

1+sinx , ans. Is π2 (π−2) ]

Question 33: [ ∫1

( x−1 )dx +

−(x−2)dx+∫2

(x−2) dx+∫1

−(x−3) dx +∫3

(x−3) 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 ∫0

e2−3xdx as a limit of sum.

[nh =1, ∫0

e2−3x =limh→0 h¿¿+………+f((n-1)h)]limh→0

h¿¿+…………..e2−3 (n−1)h ]e2 lim

h→0h. e

−3−1e−3h−1

( ∵ nh=1) =(e−3−e2 ¿ .−13 (limh→ 0

eh−1h

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 – 2Question 1:Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4yQuestion 2:Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y 2 = 1Question 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 isA. 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 evaluateQuestion 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 = x2 if x > 0 and y = –x2 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

Integration Problems

1. ∫(2x3 + 5x + 1)e2x dx [ans. e2x(x3 – 3/2x2 + 4x – 3/2) + C ]2. ∫cos2 x tan2 x dx [ x/2− 1/4 sin 2x + C]3. ∫ecos 2x sin x cos x dx [ (−1/4) ecos 2x + C]

4.∫ sec ² x2+ tan x dx [ ln |2 + tan x| + C]

5. ∫ x ²−1x+3

dx [ x2/2 − 3x + 8 ln |x + 3|

6. ∫ x3

√4+x ² dx [ 1/3(x2 + 4)3/2 − 4(x2 +

4)1/2 + C]

7. ∫ 1√ x (1+3 √x)dx [ 2/3 ln |1 + 3√x| +

8. ∫ x3+2x ²+3 x+4x3+4 x

dx [x +ln|x|+1/2ln|x2+4|

−1/2arctan(x/2)+c]9. ∫ ( sinx+cosx ) ² sinx dx [2/3 sin3x – cosx+c]

10. ∫0

secx tan3 x dx [ (2 - √2)/3 ]

**Question: Integrate . Let

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

dx = (2u-8) du .

In addition, the range of x-values is

so that the range of u-values is

INTEGRATION AS LIMIT OF A SUM AND ITS EVALUATION

Question: Divide the interval into equal parts each of length

SOLUTION : Divide the interval into equal parts each of length

Question: Divide the interval into equal parts each of length

(Recall that .)

(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.)

**Question: Integrate . Note that

and so that

. Therefore,

**SOLUTION : Integrate . First, use polynomial

division to divide by . The result is

. In the third integral, use u-substitution. Let

so that ,

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

. so that

. In addition, we can "back substitute" with

**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

**Question: Integrate . Factor and decompose into partial fractions (There is a repeated linear factor !), getting

**SOLUTION : Integrate . Factor and decompose into partial fractions (There are two repeated linear factors !), getting

**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.)

it follows that and and ;

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

**Question: Integrate . Because we want to simultaneously eliminate a square root and a cube root, use the power substitution

so that

(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.)

**SOLUTION : Integrate . Remove the cube root first. Use the power substitution

so that

**Question: Integrate . Remove the ``outside" square root first. Use the power substitution

so that

**SOLUTION : Integrate . Use the power substitution

so that

Use the method of partial fractions. Factor and decompose into partial fractions, getting (There are repeated linear factors!)

(Recall that .)

**SOLUTION : Integrate . Use the power substitution

so that

Use the method of partial fractions. Factor and decompose into partial fractions, getting

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 1:

Find the area between y = 7 – x2 and the x-axis between the values x = –1 and x = 2.

Example 2:

Find the net area between y = sin x and the 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 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

figure 3

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, 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.

figure 4

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:

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 unitsExample - 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 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 2. Set up an integral.

Step 3. Evaluate the integral.

ASSIGNMENT OF INTEGRATION

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

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

(iii) ∫0

secx tan3 x dx [answer is (2 - √2)/3 ]

(iv) ∫ dx[multiply&divide by sin(a-b)](v)∫ √ 1−√ x1+√x dx

[multiply & divide by √1−√ x ] (Vi)∫ x

x3−1dx [by partial fraction]

(v)∫ (x−4)ex

(x−2)3 dx [ use ∫ex(f(x)+f’(x))dx] (vi)∫

π /2dx

3+2 sinx+cosx dx [put sinx=

2 tanx /21+ tan ² x /2, cosx =1−tan ² x /2

1+tan ² x /2 , then put t=tanx/2. Answer is tan−12– п/2]

(vii) ∫0

¿ xcosπx∨¿¿ dx [∫0

¿ xcosπx∨¿¿+ ∫1 /2

¿ xcosπx∨¿¿ = ∫+ve dx+∫ -ve dx ,

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

п/6] (ix) ∫0

√ tanx + √cotx dx [ answer is √2π] (x) ∫0

sin−1√ xa+x dx [ put x=atan2Ѳ

, answer is a/2( -2) ] п (xi) ∫0

1+sin ² x dx [ use property ∫0

f ( x )dx = ∫0

f (a− x )dx , ∫0

f ( x )dx =2∫0

f ( x )dx ∵f(2a-x) = f(x) , then put t=tanx, answer is /2√2 ] п² (xii) ∫

f ( x ) dx , where f(x) =|x|+|x+2|+|x+5|. [

∫−5

(−x+3)dx + ∫−2

( x+7) dx , answer is 31.5 ] (xiii) Evaluate ∫ ex (1−x ) ²

( x ²+1 ) ² dx

[use ∫ ex(f(x)+f’(x))dx Question 2 Using integration, find the area of the regions: (i) { (x,y): |x-1| ≤y ≤√5−x ² } (ii) {(x,y):0≤y≤x2+3; 0≤y≤2x+3; 0≤x≤3}[(i) A= ∫

√5−x ²dx- ∫−1

(−x+1 )dx - ∫1

(x−1) dx = 5/2 [ sin−1( 2√5

) +sin−1( 1√5

] – ] [(ii) ½ A=∫0

(x ²+3¿)¿ dx +∫2

(2 x+3)dx , answer is 50/3](iii) Find the area bounded by the curve x 2 = 4y & the line x = 4y – 2.[A = ∫

dx - ∫−1

2x ²4

dx = 9/8 sq. Unit.]**(iv) Sketch the graph of f(x) = {|x−2|+2 , x≤2x ²−2 , x>2 ,evaluate∫

f (x ) dx[hint: ∫

f (x )dx = ∫0

(4−x )dx + ∫2

(x ²−2) dx = 62/3.]

**Question 3 evaluate ∫ √1+ x√ x

dx [ mult. & divide by √1+x , 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 : ∫

f ( x )dx limh→0

∑r=1

f (a+rh),

where nh=b-a & n→∞ Question 4 Evaluate ( i )∫0

¿¿) dx (ii) ∫0

(x ²−2 x+2) dx[ use lim

h→ 0

eh−1h

= 1 for part (i) , use formulas of special sequences, answer is 6]Some special case :

(1) Evaluate: ∫ dx(x−3)√ x+1 [ put x+1=t²] (2) ∫ dx

(x ²−4)√ x+1 [ put x+1 = t² ]

(3) Evaluate: ∫ dx(x+1)√x ²−1 (4) Evaluate: ∫ dx

x ²√x ²+1 [ put x=1/t for both]

(5) Evaluate: ∫ (x ²+1)dxx4+1

[ 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 ∫ x √1+x−x ² dx [ let x=A(d/dx) ( 1+x-x²) +B]

(7) Integrating by parts evaluate ∫ x ²( xsinx+cosx ) ² = ∫ ( xsecx ) . xcosx

( xsinx+cosx )²

(8) Evaluate ∫ 11+cotxdx =∫ sinx

sinx+cosxdx [ 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]

integrals with inverse trigonometric functions

forms of x

ex dx

substitute withex

original problem

3e3x dx

question paper

following indefinite

ex dxor

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