Calc length of arc of y=ln(x) from x=1 to x=2
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So far:
Definite Integral over x=(1,2) of sqrt(1 + 1/x) dx
1/x = tan^2 t
x = 1/tan^2 t
sqrt(1+1/x) = sqrt(1+tan^2 t) = sec t
dx = -2 * tan^-3 t * sec^2 t dt
Integrate over x=(1,2): sec^3 t / tan^3 t dt
Integrate over x=(1,2): 1 / sin^3 t dt
How do I solve?
Thanks!
I will assume that your integral for the path length is correct, and just worry about the integral.
The indefinite integral of sin^-3 t dt is, according to my Table of Integrals,
(-1/3)cos t/sin^2t +(1/2)ln[sin t/(1 + cos t)]
Since you have chosen to do a substitution to the variable t, the range of integration is the t range from where x = 1 to where x = 2
When x = 1, tan t = 1 and t = pi/4 =.7854
When x = 2, tan^2t = 1/2,
tan t = 1/sqrt 2 and t = 0.6155
Evaluate the indefinite integral at those two values of t, and take the difference.
To solve the integral ∫(1 / sin^3(t)) dt over the interval t=(a,b), we can use trigonometric identities and reduction formulas.
1. Rewrite the integral using the identity (1 / sin^3(t)) = csc^3(t):
∫(csc^3(t)) dt over t=(a,b)
2. Split the integral using a trigonometric identity:
∫(csc(t) * csc^2(t)) dt over t=(a,b)
3. Apply the reduction formula for the integral of csc^n(t):
∫(csc^n(t)) dt over t=(a,b) = (-1 / (n-1)) * csc^(n-2)(t) * cot(t) + (n-2) / (n-1) * ∫(csc^(n-2)(t)) dt over t=(a,b)
Using this reduction formula:
∫(csc^3(t)) dt over t=(a,b) = (-1 / 2) * csc(t) * cot(t) + 1/2 * ∫(csc(t)) dt over t=(a,b)
4. Evaluate the integral ∫(csc(t)) dt over t=(a,b) using a substitution:
Let u = cot(t), then du = -csc^2(t) dt
Rewrite the integral in terms of u:
-1/2 * ∫(du) over u=(a',b')
Since du = -csc^2(t) dt, the limits of integration will also change to be in terms of u.
5. Evaluate the integral:
∫(du) over u=(a',b') = u(b') - u(a')
Substituting back u = cot(t):
-1/2 * [cot(b') - cot(a')]
6. Simplify the expression:
Using the identity cot(t) = cos(t) / sin(t):
-1/2 * [(cos(b') / sin(b')) - (cos(a') / sin(a'))]
7. Substitute back the original limits of integration:
-1/2 * [(cos(b) / sin(b)) - (cos(a) / sin(a))]
8. Simplify further using trigonometric identities:
-1/2 * [(cos(b) * sin(a) - sin(b) * cos(a)) / (sin(a) * sin(b))]
9. Finally, simplify the expression:
-1/2 * [sin(b - a) / (sin(a) * sin(b))]
Therefore, the length of the arc of y = ln(x) from x=1 to x=2 is given by:
-1/2 * [sin(b - a) / (sin(a) * sin(b))], where a = ln(1) = 0 and b = ln(2) ≈ 0.693