29

INTEGRATION OF INVERSE TRIGONOMETRY

1) Solve the integral ∫ ex

1+e2 x dx

Solution:

Step 1: Simplify the question

∫ ex

1+e2 x dx = ∫ ex

(1)2+(ex )2dx

Step 2: Find ‘u’ , ‘a’ and dudx

→ dx = ?

u = exdudx

= ex dx = du

ee a =1

Step 3: Replace ‘u’ , ‘a’ and ‘dx’ into Step 1 and simplify.

∫ ex

(1)2+(ex )2dx = ∫ ex

(a)2+ (u )2⦁ du

ex

= ∫ du

(a)2+ (u )2= ∫ 1

(a)2+ (u )2du

Step 4: Refer to the formula sheet and compare.

∫ 1

(a)2+ (u )2du→

1a

tan−1 (u )+c

a =1 , u = ex = 11

tan−1( ex

1 )+c

2) Integrate ∫ dx

x √9 x2−1

Step 1: ∫ dx

x √9 x2−1 =

∫ 1

x √(3 x )2−(1)2dx

Step 2: `u = 3xdudx

= 3 dx = du3

a =1 x = 3u

Step 3: ∫ 1

x √(3 x )2−(1)2dx =

∫ 13u

√(u)2−(a)2⦁ du

3

= ∫ 1

u √(u)2−(a)2

3

⦁ du3

= ∫ 1

u√(u)2−(a)2du

Step 4:

∫ 1

u√u2−a2du→

1a

sec−1(ua )+c

a =1 , u = 3x = 11

sec−1( 3 x1 )+c

3) Integrate ∫ 1

√4−9 s2ds

Step 1: ∫ 1

√4−9 s2ds = ∫ 1

√ (2 )2−¿¿¿¿

Step 2: `u = 3sduds

= 3 ds = du3

a =2

Step 3: ∫ 1

√ (2 )2−¿¿¿¿ = ∫ 1

√ (a )2−¿¿¿¿

= 13∫

1

√( a )2−¿¿¿¿

Step 4:

∫ 1

√ (a )2−¿¿¿¿

a = 2 , u = 3s = sin−1( 3 s2 )+c

∴ 13

sin−1( 3 s2 )+c

∫ du

√a2−u2=sin−1 u

a+C

∫ −du

√a2−u2=cos−1 u

a+C

∫ du

a2+u2=1

atan−1 u

a+C

∫ −du

a2+u2=1

acot−1 u

a+C

∫ du

u√u2−a2=1

asec−1|u

a|+C

∫ −du

u√u2−a2=1

acosec−1|u

a|+C

EXERCISE

Solve the following:

a) ∫0

1x

√4−3 x4dx

b) ∫ 1

√1−4 x2dx

c) ∫0

21

2+9 x2 dx

29

INTEGRATION OF INVERSE HYPERBOLIC

1) Solve the integral ∫ dx

√16 x2−9

Solution:

Step 1: Simplify the question

∫ dx

√16 x2−9 = ∫ dx

√(4 x)2− (3 )2

Step 2: Find ‘u’ , ‘a’ and dudx

→ dx = ?

u = 4xdudx

= 4 dx = du4

a =3

Step 3: Replace ‘u’ , ‘a’ and ‘dx’ into Step 1 and simplify.

∫ dx

√(4 x)2− (3 )2 = ∫ 1

√(u)2−( a )2⦁ du

4

= 14∫

1

√(u)2−(a )2du

Step 4: Refer to the formula sheet and compare.

∫ 1

√(u)2−( a )2du→cosh−1( u

a )+c

a = 3 , u = 4x = cosh−1( 4 x3 )+c

∴ 14

cosh−1( 4 x3 )+c

2) Integrate ∫0

21

2+9 x2 dx

Step 1: ∫0

21

2+9 x2 dx = ∫0

21

(√2)2+(3 x)2

dx

Step 2: u = 3xdudx

= 3 dx = du3

a =√2

Step 3: ∫0

21

(√2)2+(3 x)2

dx = ∫0

21

(a)2+(u)2 •du3

= 13∫0

21

(a )2+(u )2du

Step 4:

∫ 1

(a )2+(u )2du →

1a

tan−1( ua )+c

a =√2 , u = 3x = 13 [ 1

√2tan−1( 3 x

√2 )]0

2

∫ du

√a2+u2=sinh−1 u

a+C

∫ du

√u2−a2=cosh−1 u

a+C

∫ du

a2−u2=1

atanh−1 u

a+C

∫ du

u2−a2=−1

acoth −1 u

a+C

∫ du

u√a2+u2=−1

acosech −1|u

a|+C

∫ du

u√a2−u2=−1

asech −1 u

a+C

27

29

= 1

3√2 [ tan−1(3 (2)√2 )−tan−1( 3(0)

√2 )]❑

❑

3) Integrate ∫ dx

x √1−9 x2

Step 1: ∫ dx

x √1−9 x2 = ∫ dx

x √(1 )2−¿¿¿¿

Step 2: u = 3xdudx

= 3 dx = du3

a =1 x = u3

Step 3: ∫ dx

x √(1 )2−¿¿¿¿ = ∫

1u3√ (a )2−¿¿¿

¿

= ∫ 3

u√ (a )2−¿¿¿¿

Step 4:

∫ 1

√ (a )2−¿¿¿¿

a = 1 , u = 3x = sec−1 3 x+c

INTEGRATION OF INVERSE HYPERBOLIC/INVERSE TRIGONOMETRIC (COMPLETING THE SQUARE)

1) Solve the integral ∫ dx

√ x2+4 x+5

Solution:

Step 1: Simplify the question

∫ dx

√ x2+4 x+5

∫ dx

√ x2+4 x+5 = ∫ dx

√(x+2)2+(1)2

Step 2: Find ‘u’ , ‘a’ and dudx

→ dx = ?

u = x + 2 dudx

= 1 dx = du a =1

Step 3: Replace ‘u’ , ‘a’ and ‘dx’ into Step 1 and simplify.

∫ dx

√(x+2)2+(1)2 = ∫ 1

√(u)2+(a )2du

Step 4: Refer to the formula sheet and compare.

∫ 1

√(u)2+(a )2du →sinh−1( u

a )+c

a = 1 , u = x + 2 = sinh−1( x+21 )+c

2) Integrate ∫72

51

√4 x2−16 x+7dx

Step 1: ∫72

51

√4 x2−16 x+7dx

EXERCISE

Solve the following:

e) ∫ 1

x √( ln x )2−1dx

f) ∫6

9dx

√ x2+9

g) ∫ dx

√4 x2−36

x2 + 4x + 5

= (x + 2)2

(x + 2) (x + 2)

= x2 + 2x + 2x +4

= x2 + 4x +4

5 – w = 4 w = 1

∴ ( x+2 )2+1

4x2 - 16x + 7

= 4 (x2−4 x+ 74 )

(x - 2)2

(x - 2) (x - 2)

= x2 - 2x - 2x +4

= x2 - 4x +4

74

– w = 4

w = −94

∴4 [( x−2 )2−94 ]

28

29

∫72

51

√4 x2−16 x+7dx =

∫72

51

√4 [ ( x−2 )2−94 ]

dx

=

1√4

∫72

51

√(x−2)2−(√ 94 )

2dx

=

12∫

72

51

√(x−2)2−( 32 )

2dx

Step 2: u = x – 2 dudx

= 1 dx = du a = 32

Step 3:

12∫

72

51

√(x−2)2−( 32 )

2dx

=

12∫

72

51

√(u)2−(a )2du

Step 4:

∫72

51

√(u)2−( a )2du →[cosh−1( u

a )]72

5

a = 32

, u = x – 2 =[cosh−1( x−232 )]7

2

5

=12 [cosh−1 2(x−2)

3 ]72

5

∴ 12 [cosh−1 2(5−2)

3−cosh−1

2(72−2)

3 ]=0.6585❑

❑

INTEGRATION OF HYPERBOLIC FUNCTIONS

1) Solve the integral ∫ (2 x−1 ) sinh ( x2−x ) dx

Solution:

Step 1: Simplify the question

∫ (2 x−1 ) sinh ( x2−x ) dx

Step 2: Find ‘u’ , ‘a’ and dudx

→ dx = ?

u = x2−xdudx

= 2x - 1 dx = du

2 x−1

Step 3: Replace ‘u’ and ‘dx’ into Step 1 and simplify.

∫ (2 x−1 ) sinh ( x2−x ) dx =

∫ (2 x−1 ) sinh u •du

2 x−1

= ∫sinh u du

Step 4: Refer to the formula sheet and compare.

∫sinh u du → cosh❑u+c

EXERCISE

Solve the following:

h) ∫ dx

√2 x2−4 x+5

i)∫ dx

√ x2−3x+ 52

∫cosh u du = sinh u + c

∫sinh u du = cosh u + c

∫sech2 u du = tanh u + c

∫cosech2 u du =−coth u + c

∫sech u tanh u du =−sech u + c

∫cosech u coth u du =−cosech u+c

29

u = x2−x = cosh❑(x2−x )+c

2) Integrate ∫1

2

cosh❑2 x dx

Step 1: ∫1

2

cosh❑2 x dx

Step 2: u = 2xdudx

= 2 dx = du2

Step 3: ∫1

2

cosh❑2 x dx = ∫1

2

cosh❑u •du2

= 12∫

1

2

cosh❑u du

Step 4:

∫1

2

cosh❑u du → [sinh❑u ]12

u = 2x = [sinh❑2x ]12

= 12

[sinh❑2 x ]12

∴ 12

[sinh❑2 (2 )−sinh 2(1)]❑❑=11.832

3) Integrate ∫ tanh 10 x dx

Step 1: ∫ tanh 10 x dx = ∫ sinh 10 xcosh 10 x

dx

Step 2: u = cosh 10 x dudx

= 10 sinh 10x dx =

du10 sinh10 x

Step 3: ∫ sinh 10 xcosh 10 x

dx = ∫ sinh 10 xu

•du

10 sinh10 x

= 1

10∫1u

du

Step 4: Integrate

110∫

1u

du →1

10ln u+c

u = cosh 10 x = 1

10ln ¿¿

EXERCISE

Solve the following:

h) ∫ dx

√2 x2−4 x+5

i)∫ dx

√ x2−3x+ 52

EXERCISE

Solve the following:

l) ∫ cosh x

√8+ cosh2 xdx

m) ∫12 x2 sech2 (4 x3 ) dx

n) ∫sinh 2 x dx