Mathematics in Simplified Way

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9 1

67

     

(

)

du dx d csc −1 u = − dx u u2 −1

6

«u «> 1

Example: y = sin −1 x 2 y′ =

1

.

( )

2

1− x2

( )

d 2 2x x = dx 1− x4

Example: y = tan −1 x + 1 y′ =

1+

1

(

x +1

)

2

.

d dx

(

)

x +1 =

1

(x + 2)

.

1 2 x +1

=

1 2 x + 1.( x + 2 )

Example: y = sec −1 (3 x ) d 3 1 = (3x ) = 2 dx 3x 9 x − 1 x 9 x 2 − 1 −1

1

y′ =

(3x )

3x

4.4

.

2

Integration Formula:

du

1

³

2

³

3

³1+ u

4

³1+ u

5

³u

u 2 −1

6

³u

u 2 −1

1− u2 du 1− u2 du

= sin −1 u + C = − cos −1 u + C

2

= tan −1 u + C

2

= − cot −1 u + C

du

u2 < 1

du

du

For all u

= sec −1 u + C = − csc −1 u + C

68

     

1

[

dx

³1+ x

Example:

]

1

= tan −1 x 0 = tan −1 1 − tan −1 0 =

2

0

2

³

Example:

2

dx x x2 −1

[

= sec −1 x

]

2

=

2 3

π 4

−

π 6

=

π 4

−0 =

π 4

π 12

3

Note: angle = θ = sec −1 ∴ sec −1

2

2

= 30 0 =

3

3

2

Ÿ∴ sec θ =

3

=

1 cos θ

π

1

θ

2

6 √3

³

Example: Evaluate Solution:

³

x

2

1− x6

dx

Compare with

du

= sin −1 u + C 1− u2 ∴ u 2 = x 6 Ÿ u = x 3 Ÿ du = 3 x 2 dx ∴³

x2 1− x

6

dx =

1 3 x 2 dx ³ 3 1 − x3

Example: Evaluate

( )

2

=

³

( )

1 1 1 du = sin −1 u + C = sin −1 x 3 + C 3 ³ 1− u2 3 3

dx 9 − x2

Solution: dx dx =³ 2 § x2 · § x· 9¨¨1 − ¸¸ 3 1− ¨ ¸ 9 ¹ ©3¹ © x 1 ∴ u = Ÿ du = dx, or , dx = 3du 3 3 dx du 3du § x· ∴³ =³ =³ = sin −1 u + C = sin −1 ¨ ¸ + C 2 2 2 ©3¹ 9− x 3 1− u 1− u

³

dx

9 − x2

=³

69

     

4.5

The Natural Logarithm:

The natural logarithm y=ln x d (ln u ) = 1 du dx u

Derivative

Example: find dy/dx if y=ln(3x2+4) Solution:

dy 1 = (6 x ) = 62 x dx 3 x 2 + 4 3x + 4

³

And, integration is

du = ln u + C u

Example: π

cos θ

2

³π 2 + sin θ dθ

−

2

∴ u = 2 + sin θ Ÿ du = cos θdθ π

∴

2

³π

−

π

2 cos θ du dθ = ³ = ln u + C = [ln (2 + sin θ )] π = ln (2 + 1) − ln (2 + (− 1)) 2 + sin θ u − 2

2

= ln 3 − ln 1 = ln 3 note, ln 1 = 0

Example: Evaluate

³

ln x dx x

Solution: 1 dx x (ln x )2 + C ln x u2 ∴³ +C = dx = ³ udu = x 2 2 let , u = ln x Ÿ du =

General formula for y=tanx and y=cotx:

³ tan udu = − ln cos u + C = ln sec u + C ³ cot udu = ln sin u + C

70

     

Example: Proof that ³ tan xdx = ln sec x + C ? Solution:

sin x

³ tan xdx = ³ cos x dx = − ³ = ln (cos x )

−1

(− sin x ) dx = − ln cos x + C cos x

+ C = ln sec x + C

³ 2 x tan (5 x

Example: Evaluate

2

)

−1 dx

Solution:

³ 2 x tan(5 x 4.6

2

)

− 1 dx =

(

)

(

)

1 1 tan 5 x 2 − 1 .(10 xdx ) = ln sec 5 x 2 − 1 + C 5³ 5

Properties of Natural Logarithm:

Properties of y=lnx are: 1. Domain, the set of positive real number, x>0. 2. Range, all real numbers, -∞< y