Integrals Yielding Inverse Trigonometric Functions (Tagalog/Filipino Math)
Integrals Yielding Inverse Trigonometric Functions (Tagalog/Filipino Math)

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.

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
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.
answer questions, complete that section and only then move
to short or very short answer section.
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
(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
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&divide 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