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

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SUMMARY

The discussion focuses on the differentiation and integration of functions involving inverse sine and polynomial expressions. The derivative of the function \( y = \sin^{-1} \left( \frac{X+a}{b} \right) \) is established as \( \frac{1}{\sqrt{b^2 - x^2 - 2ax - a^2}} \). The integration of \( \frac{x}{\sqrt{9-4x-x^2}} \) leads to a substitution method where \( u = 4x + x^2 \) simplifies the integral. The constants \( a \) and \( b \) are determined to be \( a = 2 \) and \( b = \sqrt{13} \) based on the equations derived from the integration process.

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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.
 
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welcome to pf!

hi lpheng! welcome to pf! :smile:

fiddle about with the a and b in your first answer until you get 9-4x-x2 :wink:
 
lpheng said:
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.
 
Last edited:
matphysik said:
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.
 

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