(1.)  dx/(4 - x2)1/2          here is the problem

Let x2 = 4 cos2 u

x = 2 cos u           take square roots

dx = -2 sin u du       take the derivative

(-2 sin u du)/(4 - 4cos2 u)1/2   make substitutions

=   -
du               simplify using trig

=   -u + C               integrateA

=   -arccos (x/2) + C      replace u with arccos (x/2)

(2.)
dx/(x2 - 25)1/2             here is the problem

Let x2 = 25 sec2 u

x =  5 sec u              take square roots

dx = 5 sec u tan u du     take the derivative

(5 sec u tan u du)/(25sec2 u - 25)1/2   make substitutions

=
sec u du       simplify using trig

=
sec u * [sec u + tan u]       multiply by this form
____________________________du  of the number "1"
[sec u + tan u]

=   ln [sec u + tan u] + C          integrate
5, (x2 - 25)1/2   ,  x     use this right triangle

=  ln [arcsec (x/5) + [(x2 - 25)1/2/5] + C

(3.)
[x/(9 - x2)1/2]dx             here is the problem

Let x2 = 9 cos2 u

x = 3 cos u

dx = -3 sin u du

[(3 cos u)(-3 sin u)du]/(9 - 9cos2 u)1/2

[make substitutions]

=  -3
cos u du         simplify using trig

=  3 sin u + C                integrate

[use this right triangle:  x, (9 - x2)1/2 , 3

=  3(9 - x2)1/2
___________ + C                use the triangle
3

=  (9 - x2)1/2 + C                     cancel the 3's

(4.)
[x/(25 - x2)3/2]dx

Let x2 = 25 cos2 u

x = 5 cos u

dx = -5 sin u du

(5 cos u * -5 sin u du)/(25 - 25cos2 u)3/2

[make substitutions]

= -(1/5)
sin-2 u cos u du    simplify using trig

=  (1/5)(sin u)-1  + C        integrate

=  (1/5)csc u + C               reciprocal id

use this right triangle:   x, (25 - x2)1/2 ,5

= (1/5)[5/(25 - x2)1/2] + C     use the triangle

=   (25 - x2)-1/2 + C          cancel the 5's, use negative exp

(5.)
(4 - x2)1/2 dx         here is the problem

Let x2 = 4 cos2 u

x = 2 cos u

dx = -2 sin u du

(-2 sin u du)(4 - 4cos2 u)1/2   make substitutions

=  -4
sin2 u du              simplify using trig

=  -2
(1 - cos 2u)du         half angle id for sin

=   -2
du + 2cos 2u du    multiply thru parentheses

=   -2u + sin 2u + C          integrate

=  -2u + 2 sin u cos u + C    double angle id for sine

use this right triangle:  x,(4 - x2)1/2 ,2

=  -2 arccos (x/2) + 2(x/2)[(4 - x2)1/2/2] + C  use the triangle

=  -2 arccos (x/2) + (x/2)(4 - x2)1/2 + C    cancel 2's

(6.)
[x2/(9 - x2)1/2]dx         here is the problem

Let x2 = 9 cos2 u

x = 3 cos u

dx = -3 sin u du

(9 cos2 u * -3 sin u du)/(9 - 9cos2 u)1/2   make substitutions

=  -9
cos2 u du     simplify using trig

=  -4.5
(1 + cos 2u)du      half angle id

=  -4.5u - 2.25 sin 2u + C      integrate

use this right triangle:   x, (9 - x2)1/2 ,3

= -4.5u - 4.5 sin u cos u + C    double angle id for sin

=  -4.5 arccos (x/3) - 4.5[(9 - x2)1/2/3](x/3) + C

[use the triangle]

=  -4.5 arccos (x/3) - (x/2)(9 - x2)1/2 + C   simplify

(7.)
[x/(x2 + 25)1/2]dx           here is the problem

=
x(x2 + 25)-1/2 dx            use a negative exponent

=  (1/2)
(2x)(x2 + 25)-1/2 dx    multiply by 1/2 and by 2

=   (x2 + 25)1/2 + C             integrate

(8.)
[x/(x2 - 4)1/2]dx

=  (x2 - 4)1/2 + C             integrate

(9.)
x3(4 - x2)1/2 dx             here is the problem

Let x2 = 4 cos2 u

x = 2 cos u

dx = -2 sin u du

(2 cos u)3(4 - 4 cos2 u)1/2 * (-2 sin u du)

[make the substitutions]

=   -32
(cos3 u)(sin2 u)du       simplify using trig

=  -32
(1 - sin2 u)(sin2 u)(cos u)du    trig id and factor

=  -32
sin2 u cos u du + 32sin4 u cos u du    multiply thru

=  -(32/3)sin3 u + 6.4 sin5 u + C    integrate

use this right triangle:  x, (4 - x2)1/2 ,2

=  -(32/3)(1/8)(4 - x2)3/2  + 6.4(1/32)(4 - x2)5/2 + C

[use the triangle][make the substitutions]

=  -(4/3)(4 - x2)3/2 + (1/5)(4 - x2)5/2 + C      multiply

(10.)
(4 - 9x2)1/2
________________ dx         here is the problem
x2

=   2
[1 - (9/4)x2]1/2
_________________________ dx     factor like this
x2

let (9/4)x2 = cos2 u

(3/2)x = cos u

(3/2)dx = - sin u du

=   2
[1 - cos2 u]1/2 * (-2/3)sin u du
_______________________________________  make substitutions
(4/9)cos2 u

=      -3
(tan2 u) du      simplify using trig

=   -3
(sec2 u - 1)du           trig id

=   -3 tan u + 3u + C     integrate

use this triangle:  3x, (4 - 9x2)1/2 , 2

= -(1/x)(4 - 9x2)1/2 + 3 arccos (3x/2) + C   make substitutions