student note - integration techniques

8/12/2019 Student Note - Integration Techniques

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Integration Techniques

I. Definition of the Integral as an Anti-derivative

If d

dxF(x) =f(x), then the indefinite integral off(x) with respect to x is f(x) =

F(x). For example, since ddx

x2 = 2x, we say that the indefinite integral of 2xis x2, i.e.,

2x dx= x2

Note:

The indefinite integral is not unique as the derivative of, for example, x2 + 3 is also 2x,so x2 + 3 is another indefinite integral of 2x. This can be expressed by writing

2x dx= x2 +Cwhere C is arbitrary constant. The constant Chas no specific value but can be givenany value based on the boundary condition of a particular problem.

The inverse relationship between differentiation and integration means that for everystatement about differentiation, we can write down a corresponding statement aboutintegration. For example,

d

dxsin x= cos x , then we have

cos x dx= sin x

When we are given a function to integrate, we have to learn to recognize a given func-tion as the derivative of another function. In other word, we have to run through allthe standard differentiation formulae in our mind until we come to one which fits ourproblem.

II. Integrating the Power ofx

We now look at how to integrate the power ofx by looking at the corresponding rulefor differentiation:

d

dx(xn) =nxn1 , so

nxn

1

dx= xn +A ,

whereA is an arbitrary constant. Similarly,

d

dx

xn+1

n+ 1

=

1

n+ 1

d

dxxn+1 =

1

n+ 1 (n+ 1)xn =xn ,

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so that xn dx=

xn+1

n+ 1+ A .

This formula hold for all real numbers ofn except n= 1 because the division by zerodoes not make sense. When n = 1, x

n dx becomes x1 dx=

1

xdx. Since

ddx

(ln x) = 1x

, we have

1x

dx = ln x+A .

Please note that the integration such as sin2 x dx =

sin3 x

3 +c ,

(2x+ 1)6 dx = (2x+ 1)7

7 +c ,

are wrong. You will find that the differentiation of the function on right-hand side doesnot produce the integrand on the left-hand side. It is a good practice to always check

your result this way during the examination. The table of the integrals of some commonfunctions are attached in the Appendix.

III. Properties of Definite Integrals

Assume the functions f and g are continuous function in [a, b].

Property 1: ba

cf(x) dx= c

ba

f(x) dx , for any constant c.

Example: 2

1

5x dx = 5

2

1

x dx

= 5

2

2

1

2x dx

= 5

2

x221

= 5

2(2)2 (1)2

=

15

2

Example:

3/2

1/2

61 t2 dt = 6

arcsin t

3/21/2

= 6

3

6

=

2


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Example: 5

x

x2 3 dx = 5

1

xx2 (

3)2dx

= 5

3arcsec

x

3

Property 2: ba

f(x) g(x) dx= b

a

f(x) dx ba

g(x) dx

This property says that the integral of a sum (or minus) is the sum (or minus) of the

integrals.Example:

1

0

4 + 3x2

dx =

1

0

4 dx+

1

0

3x2 dx

=

4x10

+

x310

= 4 + 1 = 5

This property can be extended to the case where there are more than two terms in the

sum/difference.

Property 3:Ifa < b < c and iffintegrable on [a, b] and [b, c] thenfis integrable on [a, c] and

ca

f(x) dx=

ba

f(x) dx+

cb

f(x) dx

Example:

Integrate 53|x 2| dx

Solution:

Since

|x 2| =

x 2 if x 2(x 2) if x


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5

3|x 2| dx =

2

3(x 2) dx+

5

2

(x 2) dx

=

2

3(x+ 2) dx+

5

2

(x 2) dx

=x

2

2 + 2x23 +

x22 2x

52

= (2 + 4) (92 6) + (25

2 10) (2 4)

= 2 + 212

+ 52

+ 2 = 17

Property 4:We define the definite integral

ba

f(x) dxby assuming thata < b. If we reversea and b,then

b

a

f(x) dx=

a

b

f(x) dx

Property 5:Ifa = b, then a

a

f(x) dx= 0

Exercise:

Evaluate the integral below:

1. 20

x2

dx

2.

0

cos x dx

3.

2

2(4 x2) dx

4.

(1 t)(2 t2) dt

5.

u2 + 1 + 1u2

du

6.

(2x)2 dx

7.

(sin + 3 cos ) d

8. sin2x

sin x dx

9.

sin x

1 sin2 xdx

10.

x

x dx

11.

32

0

| sin x| dx

12. 902t dt

13.

x( 3

x+ 4

x) dx

14.

1 + cos2

cos2 d

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IV. Substitution Method

In the previous chapter we learnt how to differentiate the composition of two functionsby using the Chain Rule. For example,

d

dx

sin(x2 + 1)

by letting u = x2 + 1, we havedu

dx= 2x

thend

dxsin(x2 + 1) =

du

dx d

dusin(u) = 2x cos(u) = 2x cos(x2 + 1) .

The inverse relationship between differentiation and integration means that

2x cos(x2 + 1) dx= sin(x2 + 1) +c .

This tells us if there are product of two functions in the integrand such that one of thefunction is the derivative of the composite of another function, then

dg(x)

dx f(g(x)) dx=F(g(x)) +c

whered

duF(u) =f(u) F(u) =

f(u) du .

Theorem:

Ifu= g(x) is a differentiable function whose range is an interval Iand fis continuouson I, then

f

g(x)

g(x) dx=

f(u) du +c

In particular, wheng(x) =ax+b where aandb are constant, then f(ax+b) (ax+b) dx =

f(u) du +c

f(ax+b) a dx = f(u) du +c f(ax+b) dx =

f(u) du

a +C

An example of this particular case is illustrated below: cos(2x+ 3) dx=

sin(2x+ 3)

2 +c .

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Here f(ax+b) = cos(2x+ 3) so that a= 2, then f(u) = cos(u) and thus

f(u) du=cos(u) du= sin(u) = sin(2x+ 3).

Example:

Integrate x2 cos(x3 + 3) dx .

Solution:

Since x2 cos(x3 + 3) dx=

cos(x3 + 3) x2 dx

Let u = x3 + 3 du = 3x2 dx 13du = x2 dx, substitute this new variable into the

integrand yields

cos(x3 + 3) x2 dx = cos(u)

1

3du

= 1

3

cos(u) du

= 1

3sin(u) +c

= 1

3sin(x3 + 3) +c

Example:

Integrate

(7x+ 1)

20

dx

Solution:

Let u = 7x+ 1 thendu= 7 dx 17

du= dx. Substitute these into the integrand yields (7x+ 1)20 dx =

(u)20

1

7du

= 1

7

u20 du

= 1

7u

21

21 +c

= (7x+ 1)21

147 +c

Example:

Integrate 2x ex

2

dx

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Solution:

Let u = x2 then du = 2x dx, therefore 2x ex

2

dx=

eu du= eu +c= ex

2

+c

Example:

Integrate 2x

x2 + 5dx

Solution:

Let u = x2 + 5, then du= 2x dx, therefore 2x

x2 + 5dx =

1

udu = ln |u| +c= ln |x2 + 5|+c

Example:

Integrate 2x+ 1 dx

Solution:

Let u = 2x+ 1, thendu= 2 dx dx= 12du, therefore

2x+ 1 dx =

u 12

du

= 1

2

u du

= 1

2

u1/2 du

= 1

2u

3/2

3/2 +c

= u3/2

3 +c

= (2x+ 1)3/2

3

+c

Example:

Integrate x5

1 +x2 dx

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Solution:

Let u = 1 +x2 thendu= 2x dx x dx= 12du, and also

u = 1 +x2

u 1 = x2

(u 1)2

= x4

Substitutexwith the variable u in the integrand yields x5

1 +x2 dx =

x4

1 +x2 x dx

=

(u 1)2 u 1

2du

= 1

2

(u 1)2 (u)1/2 du

=

1

2

(u2

2u+ 1) (u)1/2

du

= 1

2

(u5/2 2u3/2 +u1/2) du

= 1

2

u7/2

7/2 2u

5/2

5/2 +

u3/2

3/2

+c

= u7/2

7 2u

5/2

5 +

u3/2

3 +c

= (1 +x2)7/2

7 2(1 +x

2)5/2

5 +

(1 +x2)3/2

3 +c

Example:

Integrate x+ 9

x2 + 9dx

Solution:

Let u = x2 + 9 thendu= 2x dx, therefore

x+ 9x2 + 9

dx = xx2 + 9

dx+ 9x2 + 9

dx

= 1

2

1

udu+ 9

1

x2 + 32dx

= 1

2ln |u|+ 9 1

3arctan

x3

+c

= 1

2ln |x2 + 9| + 3 arctan

x3

+c

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Exercise:

1. dx5 3x

2.

(ln x)2

x dx

3.

dx

x ln x

4.

sec2 x tan x dx

5. 2

1

e1/x

x2

dx

6.

1

0

x2(1 + 2x3)5 dx

7.

sec3 x tan x dx

8.

(x2 + 1)

2x3 + 6x dx

9. dx1 + 9x

2

10.

2x e3x

2

e3x2 + 2dx

11.

1

3 +x2dx

12.

x

1 4x2 dx

13. xx2

+ 9dx

14.

tan x dx

15.

sec x dx

16.

cos

t

tdt

V. Trigonometric Integrals

Objective: At the end of this topic you should be able to integrate trigonometry ofhigher orders with odd and even powers.

Methods to be used

1. In evaluating the integral of the form sinm x cosn x d x ,

(a) Ifn (the power of cosine) is odd and m is arbitrary integer,- save one cosine factor,

- for the remaining cosine factors (which has even power), use the identitycos2 x= 1 sin2 x to express the cosine in terms of sine,

- then substituteu= sin x

(b) Ifm (the power of sine) is odd and n is arbitrary integer,

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- save one sine factor,

- for the remaining sine factors (which has even power), use the identitysin2 x= 1 cos2 x to express the sine in terms of cosine,

- then substituteu= cos x

Example: cos3 x dx =

cos2 x cos x dx

=

(1 sin2 x)cos x dx

Letu = sin xthedu = cos x dx, therefore (1 sin2 x)cos x dx =

(1 u2) du

= u

u3

3

+c

= sin xsin3 x

3 +c

Example: sin5 x cos2 x dx =

sin4 x sin x cos2 x dx

=

(sin2 x)2 sin x cos2 x dx

=

(1 cos2

x)2

sin x cos2

x dx

Letu = cos xthendu= sin x dx, therefore (1 cos2 x)2 sin x cos2 x dx

=

(1 u2)2 u2 (du)

=

(1 2u2 +u4)u2 du

= (u2 2u4 +u6) du=

u3

32u

5

5 +

u7

7

+c

= cos3 x

3 +

2 cos5 x

5 cos

7 x

7 +c

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(c) Ifm andnare both even, then use half angle identities

sin2 x = 12

(1 cos2x)cos2 x = 1

2(1 + cos 2x)

sin x cos x = 12sin 2x

In general,

sin2 nx = 12

(1 cos2nx)cos2 nx = 1

2(1 + cos 2nx)

sin nx cos nx = 12sin 2nx

For example,

sin2 2x = 12

(1 cos4x)cos2 3x = 1

2(1 + cos 6x)

sin4x cos4x = 12sin 8x

Example: sin4 x dx =

(sin2 x)2 dx

=

1 cos2x

2

2dx

=

1

4(1 2cos2x+ cos2 2x) dx

= 14

1 2cos2x+12 (1 + cos 4x)

dx

= 1

4

3

2x 2 sin 2x

2 +

sin 4x

2.4

+c

= 3

8x sin 2x

4 +

sin 4x

32 +c

2. In evaluating

tanm x secn x dx(a) Ifn is even andm is arbitrary integer,

- save a factor of sec2 x

- substitute sec2 x= 1 + tan2 x

- then substituteu= tan x

(b) Ifm is odd and n is arbitrary integer,

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- save the factor of sec x tan x

- substitute tan2 x= sec2 x 1- then substituteu= sec x

Example:

tan3 x dx =

tan2 x tan x dx

=

(sec2 x 1)tan x dx

=

sec2 x tan x sin x

cos x

dx

For the first term, let u= tan x then du= sec2 x dx; for the second term, letv= cos x thendv= sin x. Therefore,

sec2 x tan x dx sin x

cos xdx = u du+ dv

v

= u2

2 + ln |v| +c

= tan2 x

2 + ln | cos x| +c

Example: sec4 x tan3 x dx =

sec3 x tan2 x(sec x tan x) dx

=

sec3 x(sec2 x 1)(sec x tan x) dx

Letu = sec x, then du = sec x tan x dx, therefore sec3 x(sec2 x 1)(sec x tan x) dx

=

u3(u2 1) du

=

(u5 u3) du

= u6

6u

4

4 +c

= sec6 x

6 sec

4 x

4 +c

3. To evaluate the integrals

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a)

sin mx cos nxdx

b)

sin mx sin nxdx

c) cos mx cos nxdxwe use the identity

(i) sin A cos B= 12

[sin(A B) + sin(A+B)](ii) sin A sin B= 1

2[cos(A B) cos(A+B)]

(iii) cos A cos B= 12

[cos(A B) + cos(A+B)]Example:

sin4x cos5x x = 1

2 [sin(x) + sin 9x] dx=

1

2

[ sin(x) + sin 9x] dx

= 1

2

cos x 1

9cos 9x

+c

Exercise:

Evaluate the integrals

1.

cos2 x tan3 x dx

2.

sin5x sin2x dx

3. /40

sin4 x cos2 x dx

4.

0

sin4(3t) dt

VI. Integration by Parts

Recall from the previous lecture that the Product Rule is applied when we differen-tiate the product of some functions, i.e.,

(u(x) w(x))= u(x) w(x) +u(x) w(x) ,where denotes the derivative with respect to x. Now, if we integrate both sides of theequation above over x gives

u(x) w(x) =

u(x) w(x) +u(x) w(x) dx=

u(x) w(x) dx+

u(x) w(x) dx

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Let v(x) = w (x) so that w(x) =

v(x) dx, the above equation becomes

u(x)

v(x) dx=

u(x)

v(x) dx

dx+

u(x) v(x) dx

Thus, rearranging terms u(x) v(x) dx= u(x)

v(x) dx

u(x)

v(x) dx

dx

This formula helps us to transform the integral of products of functions into other (ideallysimpler) integrals. If an interval with endpoints aand b are given,

ba

u(x)v(x) dx=

u(x)

v(x) dx

ba ba

u(x)

v(x) dx

dx

Given a single function to integrate, the typical strategy is to carefully separate it into

a product of two functions u(x)v(x) such that the integral produced by the integrationby parts formula is easier to evaluate than the original one. The best strategy to take iseither choosing u(x) as a function that simplifies when differentiated, or choosing v(x)as a function that simplifies when integrated.

Example:

Given

x sin x dx.

since the derivative ofx is 1, we make this part as u(x) and let v= sin x, then

x sin x dx = x

sin x dx (x) sin x dx dx

= x( cos x)

(1)( cos x) dx

= x cos x+

cos x dx

= x cos x+ sin x+c

Note that if we choose u(x) = sin x and v(x) = x, this will lead to more complicatedintegral and we are unable to evaluate the integration any further, i.e.,

x sin x dx = sin x

x dx (sin x) x dx dx

= x2

2 sin x 1

2

x2 cos x d x .

It is very important to choose the u(x) and v(x) in such a way that a much simplerintegral will be obtained.

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Example:

Integrate

ln x dx.

Solution:

Let u(x) = ln x and v(x) = 1, therefore ln x dx = ln x

(1) dx

(ln x)

(1)dx

dx

= x ln x

1

x x dx

= x ln x

(1)dx

= x ln x x+c

Example:

Integrate

t2et dt.

Solution:

Let u = t2 andv = et, then t2et dt = t2et

2t et dt

= t2et 2

tet dt

For the second term on the right,

tet dt, we apply the integration by parts again bylettingu= t andv = et, then

tet dt= tet

(1)et dt= tet et

Thus, t2et dt= t2et 2(tet et) +c

Exercise:Evaluate the integrals

1.

x4(ln x)2 dx

2.

0

t sin3t dt

3.

2

1

ln x

x2 dx

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4.

x cos5x dx

5.

cos(ln x) dx

6.

e2 sin3 d

VII. Integration of Rational Functions Using Partial FractionObjective: In this topic, we introduce how partial fraction expansion provides an

approach to integrating a general rational function.

IfP(x) andQ(x) are two polynomials in x, then the ratio of these two polynomials

P(x)

Q(x) where Q(x) = 0

is called a rational function. If the degree of the numerator of the rational function is lessthan that of the denominator, the rational function is called a proper rational function.

Any proper rational function can be expressed as sum of Partial fraction.

Example:

Convert the proper rational function

x+ 5

x2 +x 2in term of partial fraction.

Solution:

We first factorize the denominator as

x+ 5

x2 +x 2= x+ 5

(x 1)(x+ 2) .Let

x+ 5

(x 1)(x+ 2) = A

x 1+ B

x+ 2

= A(x+ 2) +B(x 1)

(x 1)(x+ 2)By equating the numerator part, we have

x+ 5 =A(x+ 2) +B(x 1)There are two methods to solve for the constant Aand B .

1. Using substitution ofx value:Let x = 1, then

6 = A(3) +B(0)

A = 2

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Let x = 2, then

3 = A(0) +B(3)B = 1

2. Comparing the coefficients:

x+ 5 = A(x+ 2) +B(x 1)= (A+B)x+ 2A B

Compare the term with x:1 =A+B (1)

Compare the term with constant:

5 = 2A B (2)

From Eq. (1),A= 1 B and substitute this into Eq. (2) gives

5 = 2(1 B) BB = 1

So thatA= 1 B = 2

The result isx+ 5

x2 +x

2=

2

x

1 1

x+ 2

From this result, the integral of the rational function can be evaluate as x+ 5

x2 +x 2dx =

2

x 1 1

x+ 2

dx

= 2 ln |x 1| ln |x+ 2| +c

If the degree of the numerator is equal or greater than the degree of the denominatorin a rational fraction, then the rational function is called improper rational function.The improper rational function can be written as the sum of a polynomial function anda finite number of proper rational fractions and this proper rational fractions can beresolved into partial fractions. For instance,

4x3 3x+ 5x2 2x = 4x+ 8 +

13x+ 5

x2 2x= 4x+ 8 5

2x+

31

2(x 2)

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Let x = 0,0 =A(1)(1) + 2(1) 1(1)2 , A = 1

Thus, the integral becomes

4x(x 1)2(x+ 1)

dx

=

1

x 1+ 2

(x 1)2 1

x+ 1

dx

= ln |x 1| ln |x+ 1| + 2

1

(x 1)2dx

For the integral in the last term, let u = x 1, then du = dx, so that 1

(x 1)2dx =

1

u2du

=

u1

= 1u

= 1x 1

The result is 4x

(x 1)2(x+ 1) dx = ln |x 1| ln |x+ 1| 2

x 1+ c

Example:Evaluate the integral

x+ 4

x2 + 2x+ 5dx

Solution:

Since the denominator of integrand cannot be factorized further, this rational functioncannot be converted into the partial fraction as before. Now, let

u = x2 + 2x+ 5

du = (2x+ 2) dx

= 2(x+ 1) dxdu

2 = (x+ 1) dx

Then x+ 4

x2 + 2x+ 5dx =

(x+ 1)

x2 + 2x+ 5dx+

3

x2 + 2x+ 5dx

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For the first term on the right hand side, by the substitution in above (x+ 1) dx

x2 + 2x+ 5 =

1

2

du

u

= 1

2ln |u|

= 12

ln |x2 + 2x+ 5|For the second term on the right hand side,

3

x2 + 2x+ 5dx = 3

1

(x+ 1)2 1 + 5 dx

= 3

1

(x+ 1)2 22dx

= 3

2arctan

x+ 12

The result is x+ 4

x2 + 2x+ 5dx=

1

2ln |x2 + 2x+ 5|+3

2arctan

x+ 12

+c

Example:

Evaluate the integral x3 +x

x 1 dx

Solution:

Since the polynomial in the numerator has a degree higher than the denominator, wecannot use the integration by partial fraction directly. In this case, we perform longdivision

x2 +x+ 2

x 1 x3 +x x3 +x2

x2 +x x2 +x

2x 2x+ 2

2Thus,

x3 +x

x 1 dx =

x2 +x+ 2 + 2

x 1dx

= x3

3 +

x2

2 + 2x+ 2ln |x 1| +c

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Since cos 0 in the interval2

2, therefore| cos | = cos .

9 x2

x2 dx =

cos2

sin2 dx

= cot2 d=

csc2 1 d

= cot +c

We must return the variable to the original variable x. From earlier x = 3 sin meansthat sin = x

3, thus = arcsin(x

3). Now, how can we express cot in terms ofx? This

can be either with the aid of the diagram or by analytic method. Let us introduce thefirst method. The diagram below shows is interpreted as an angle of a right triangle:

9 x2

x3

Since sin = x3

, so that x is labeled in the opposite side of and 3 is the length of

diagonal. Then by Pythagorean Theorem, the length of the base is 32

x2

= 9 x2

.The value of cot can be constructed from the diagram as

cot =

9 x2

x . (3)

Now we show how the analytical method will also produce the same result. We use the

identity cos2 + sin2 = 1, then cos =

1 sin2 = 1 (x3

)2, so that

cot = cos

sin

=

1 (x

3 )2

x3

=

32 x2

3 3

x

=

9 x2

x .

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And so 9 x2

x2 dx = cot +c

=

9 x2

x arcsin(

x

3

) +c

Exercise:

Evaluate the integrals

1.

1

x2

x2 + 4dx

2.

dx

x2 a2 , where a >0.

3.

33/2

0

x3

(4x2 + 9)3/2dx

4.

x

1 x4 dx

5.

x

x2 7dx

6.

dx

x2 + 16

7.

1

(x2 + 2x+ 2)2dx

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Appendix: The Integrals of some common functions

1.

ex

dx= ex + c

2.

ax

dx= a

x

ln a+ c

3.

1

xdx= ln |x| +c

4.

ln(x) dx= x ln(x) x+ c

5.

cos xdx= sin x+c

6.

sin xdx= cos x+c

7.

tan xdx= ln | cosx| +c

8.

cotxdx= ln | sin x| +c

9.

secxdx= ln | secx+ tan x| +c

10.

cscxdx= ln | cscx+ cotx| +c

11.

sec2 xdx= tan x+c

12.

csc2 xdx= cotx+c

13.

secx tan xdx= sec x+ c

14.

cscx cot xdx= cscx+ c

15.

coshxdx= sinh x+ c

16.

sinh xdx= coshx+ c

17.

tanh xdx= ln(cosh x) + c

18. cothxdx= ln | sinh x| +c

19.

sechxdx= arctan(sinh x) + c

20.

cschxdx= ln

tanh

x

2+ c

21.

sech 2xdx= tanh x+ c

22.

csch 2xdx= cothx+ c

23.

sechx tanh xdx= sech x+ c

24.

cschx cothxdx= csch x+ c

25.

1a2 x2 dx= arcsin

x

a

+c

26. 1a2 x2 dx= arccos

xa

+c

27.

1

a2 +x2dx=

1

a arctan

x

a

+ c

28.

1

xx2 a2 dx=

1

aarccsc

x

a

+ c

29.

1

xx2 a2 dx=

1

aarcsec

x

a

+ c

30.

1

a2 +x2 dx=

1

aarccot

x

a

+ c

31. 1x2 +a2 dx= arcsinh xa+ c

32.

1x2 a2 dx= arccosh

x

a

+c

33.

1

a2 x2 dx= 1

aarctanh

x

a

+c

34.

1

xx2 + a2

dx= 1

aarccsch

x

a

+ c

35.

1

xa2 x2 dx=

1

aarcsech

x

a

+ c

36. 1a2 x2

dx= 1

a

arccoth xa+c

24

student note - integration techniques

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