It might simple to you, but i'm stuck.

[lpheng]

The question is something like this, find the:
d/dx of sin^-1 (X+a) /b.
then evaluate,
intergarte of X / (9-4x-x²) ^1/2
&
intergarte of X² / (9-4x-x²) ^1/2


What i get for the first part is,
d/dx of sin^-1 (X+a) /b = 1 / ( -X²/b²-2aX/b²-a²+b²/b²)^1/2


Then I'm stuck. xD
Sorry for the confusing equation.

 

[tiny-tim]

welcome to pf!

hi lpheng! welcome to pf!

fiddle about with the a and b in your first answer until you get 9-4x-x²

 

[matphysik]

The question is something like this, find the:
d/dx of sin^-1 (X+a) /b.
then evaluate,
intergarte of X / (9-4x-x²) ^1/2
&
intergarte of X² / (9-4x-x²) ^1/2What i get for the first part is,
d/dx of sin^-1 (X+a) /b = 1 / ( -X²/b²-2aX/b²-a²+b²/b²)^1/2Then I'm stuck. xD
Sorry for the confusing equation.

With y=sin⁻¹[(x+a)/b], dy/dx= 1/√[b² - x² - 2xa - a²]

Let ℐ≡ ∫ xdx/√(9-4x-x²). Set u:= 4x+x²⇒½du-2dx=xdx. Whence ℐ= ∫ ½du/√(9-u) - 2sin⁻¹[(x+2)/√5] + constant. And so on...

NOTE1: 2a=4 ⇒ a=2 and b² - a²=9 ⇒ b=√5.

NOTE2: Given, ½du-2dx=xdx multiply through by `x` to get ½xdu-2xdx=x²dx.

 

[matphysik]

Let ℐ≡ ∫ xdx/√(9-4x-x²). Set u:= 4x+x²⇒½du-2dx=xdx. Whence ℐ= ∫ ½du/√(9-u) - 2sin⁻¹[(x+2)/√5] + constant. And so on...

NOTE1: 2a=4 ⇒ a=2 and b² - a²=9 ⇒ b=√5.

Correction: b² - a²=9 ⇒ b=√13. So that,

ℐ= ∫ ½du/√(9-u) - 2sin⁻¹[(x+2)/√13] + constant.