Find arc length of y=logx from x=1 to x=2.
dy/dx)^2=1/x^2
arc length=Int of [sqrt(1+1/x^2)]dx
=Int of [sqrt(1+x^2)/x^2]
=Int of [sqrt(1+x^2)]/x from x=1 to x=2.
How to proceed further to integrate?
substitute
x = tan u
dx = sec^2 u du
1+x^2 = sec^2 u
and you will wind up with some nice integrands involving tan u and sec u
I reached upto Int csc u sec^2 u du from u=arctan 1 to 2 but am not clear how to go further.
take a trip on over to wolframalpha and enter
integral sqrt(x+1)/x dx
and then click on the "Show Step-by-Step Solution" button (you may have to register first)
and it will show all the intricacies of the substitution.
Or, recall that sec^2 = 1+tan^2.
To integrate the expression ∫[sqrt(1+x^2)]/x dx from x=1 to x=2, you can use a substitution method.
Let's substitute u = 1 + x^2.
Then, differentiate both sides with respect to x to get du/dx = 2x.
Rearranging, we have dx = du / 2x.
Substituting this back into our integral, we get:
∫ sqrt(1 + x^2) / x dx = ∫ sqrt(u) / (2x) du
Because we have x in the denominator, we still have one more step to perform before we can substitute. We can rewrite the expression as follows:
∫ u^(-0.5) / (2x) du
Now, let's substitute with u = 1 + x^2 and dx = du / (2x).
Our integral becomes:
∫ (1 + x^2)^(-0.5) / 2 du
Now, integrating with respect to u, we can treat x as a constant because u = 1 + x^2.
∫ (1 + x^2)^(-0.5) / 2 du = ∫ (1 + x^2)^(-0.5) / 2 du
To evaluate the integral between the limits of x=1 and x=2, we substitute back for u:
∫ (1 + x^2)^(-0.5) / 2 du = ∫ (1 + x^2)^(-0.5) / 2 du evaluated from u=2 to u=5.
At this point, you can evaluate the definite integral using standard integration techniques. Once you have the result, you will have the arc length of y = log(x) between x=1 and x=2.