Evaluate (integer sign)dx/sqrt9-8x-x^2
A.sin^-1 (x + 4/5) + C
B.sin^-1 (x - 4/5) + C
C.Ln Sqrt9-8x-x^2 + C
D.Ln (-8 - 2x) + C
B or C ?
∫1/sqrt(9-8x-x^2)dx
=∫1/sqrt(5²-(x-4)²)dx
substitute u=x+4, du=dx
=∫1/sqrt(5²-u²)dx
=sin-1(u/5)+C
=sin-1((x+4)/5)+C
To evaluate the integral of the expression dx/sqrt(9-8x-x^2), we can use trigonometric substitution.
First, let's write the given expression in a different form:
dx / sqrt(9 - 8x - x^2) = dx / sqrt((-1)(x^2 + 8x - 9))
Now, we can try to complete the square in the denominator:
x^2 + 8x - 9 = (x^2 + 8x + 16) - 25 = (x + 4)^2 - 25
Substituting this back into the integral:
dx / sqrt((-1)(x^2 + 8x - 9)) = dx / sqrt((-1)((x + 4)^2 - 25))
Let's substitute (x + 4) = 5*sin(t):
Then, dx = 5*cos(t)dt
Now, we can rewrite the integral in terms of the new variable t:
(5*cos(t)) / sqrt((-1)((5*sin(t))^2 - 25))
= 5cos(t) / sqrt(-25*cos^2(t))
= 5cos(t) / (5i*sin(t))
= i*cos(t) / sin(t)
= i * cot(t)
By using the trigonometric identity cot(t) = cos(t) / sin(t), we simplified the expression to just i*cot(t).
Therefore, the integral becomes:
∫ (integer sign)i*cot(t)dt
Now, let's substitute back to find the original variable x:
cot(t) = (x + 4) / 5
⇒ t = arccot((x + 4) / 5)
Finally, we can determine the indefinite integral:
∫ dx/sqrt(9 - 8x - x^2) = ∫ (integer sign)i*cot(t)dt
= (integer sign)i*arccot((x + 4) / 5) + C
So, the correct option is A. sin^-1 (x + 4/5) + C.