(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 + 2∫cos 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 + 32∫sin⁴ 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