integral of dx/square root(x^2-64)
My answer is:
ln (x/8 + square root(x^2-64)/8) +c
is this right?
not even close :-(
let x = 8secu
dx = 8secu*tanu du
x^2-64 = 64sec^2 u - 64 = 64tan^2 u
dx/√(x^2-64) = 8*secu*tanu/8tanu = secu
since ∫secu = ln|secu+tanu|, we have
Recall that x = 8secu, so
secu = x/8
tanu = √(x^2/64 - 1) = 1/8 √(x^2-64)
∫secu = ln|secu+tanu| = ln|x/8 + 1/8√(x^2-64)| + c = ln|(x + √(x^2-64))/8| + c
Now, ln(n/8) = ln(n) - ln(8), so we can fold -ln(8) into c, giving us
ln|x + √(x^2-64)| + C
hmm. upon rereading your answer, you are in fact correct.!!
To verify if your answer is correct, we can differentiate it and check if it matches the original integrand. Let's differentiate the expression you provided using the chain rule and simplify:
d/dx [ln (x/8 + sqrt(x^2-64)/8)]
= (1/(x/8 + sqrt(x^2-64)/8)) * (d/dx) (x/8 + sqrt(x^2-64)/8)
= (1/(x/8 + sqrt(x^2-64)/8)) * (1/8 + (1/8)*(1/2)*(2x)/sqrt(x^2-64))
= 1/(x/8 + sqrt(x^2-64)/8) * (1/8 + x/(8(sqrt(x^2-64))))
Now, let's simplify this expression further:
= 1/(8((x + sqrt(x^2-64))/8)) * (1/8 + x/(8(sqrt(x^2-64))))
= (1/(x + sqrt(x^2-64))) * (1/8 + x/(8(sqrt(x^2-64))))
= (1/(x + sqrt(x^2-64)) + (x/(8(x + sqrt(x^2-64))))
So, differentiating the expression you provided does not give us dx/sqrt(x^2-64). Therefore, your original answer is incorrect.
To find the correct integral of dx/sqrt(x^2-64), we can use the trigonometric substitution method.
1. Start by substituting x = 8sec(theta). This substitution is chosen because it simplifies the expression.
dx = 8sec(theta)tan(theta)d(theta)
sqrt(x^2-64) = sqrt((8sec(theta))^2 - 64) = sqrt(64tan^2(theta)) = 8tan(theta)
Now, the integral becomes:
∫ (8sec(theta)tan(theta))/(8tan(theta)) d(theta)
2. Simplify the expression further:
∫ sec(theta) d(theta)
3. Integrate the simplified expression:
∫ sec(theta) d(theta) = ln|sec(theta) + tan(theta)| + C
4. Convert back to the original variable x using the substitution x = 8sec(theta):
ln|sec(theta) + tan(theta)| + C = ln|8sec(theta)/8 + tan(theta)/8| + C = ln|x/8 + sqrt(x^2-64)/8| + C
So, the correct integral of dx/sqrt(x^2-64) is ln|x/8 + sqrt(x^2-64)/8| + C.
Therefore, your original answer is incorrect, and the correct answer is ln|x/8 + sqrt(x^2-64)/8| + C.