intergration: substitution and inverse trigonometry, help!?

Question

Substitution in intergration, help?
how to intergrate: 1/sqrt(8 - x^2 - 2*x)

i think it may involve inverse trigonometry, but just cant work it out.

i let (x^2 +2x) = u
thus: [intergral] 1 /sqrt (8 - u)

but

du/dx = 2x + 2
dx = du/ (2x+2)

there is this problem : can't substitute x back into u, thus can't intergrate the function, because in the intergral of function, there will be 2 unknown values "u" and "x".

Answer 1

First, complete the square on the bottom:

∫1/√(9-(x²+2x+1)) dx
∫1/√(9-(x+1)²) dx

Now, factor out the 9:

1/3∫1/√(1-((x+1)/3)²) dx

Now, make the substitution u=arcsin ((x+1)/3). Then 3 sin u - 1=x and 3 cos u du = dx, so:

∫1/√(1-sin² u) cos u du

Using the Pythagorean theorem:

∫cos u/√cos² u du
∫1 du

Now integrating:

u + C

And resubstituting:

arcsin ((x+1)/3) + C

And we are done.

Answer 2

ok...write 8 - x^2 - 2*x as 8 +1 -1 - x^2 - 2*x ...i.e. you add and subtract 1, so now the expression becomes 9 - (1+x)^2.
Now substitute 1+x = 3 sin u
so, dx = 3 cos u du

The integral becomes:
1/sqrt(8 - x^2 - 2*x) dx
= 1/sqrt(9 - (1+x)^2) dx
= (1/sqrt(9 - 9 sin^2 u) ) 3 cos u du
= du
= u
= arcsin (1+x)/3)

arcsin is inverse sin.....

Answer 3

familiar trigonometry applications take a level (or radians) and can provide a mode. for instance, sin (?/4)= ?2/2. The inverse trig functionality takes the fashion you gained above and can provide the perspective that corresponds to that style. So inversesin(?2/2)=?/4. that's in actuality what it truly is.

Answer 4

1/sqrt(8 - x^2 - 2*x)
=1/sqrt(9-x^2-2x-1)
=1/sqrt(9-(x+1)^2)
=1/[3sqrt(1-((x+1)/3)^2)]

Try u=(x+1)/3
du = 1/3dx

∫[1/sqrt(1-u^2)]du
= arcsin(u) + c
= arcsin[(x+1)/3] + c