prove that the integral of arcsec(x) is equal to xarcsec(x) - ln|x+ sqrt((x^2)-1)
Let x = secθ
dx = secθtanθ dθ
θ = arcsec x
Int(arcsec(x) dx)
= Int(θ secθtanθ dθ)
now integrate by parts
u = θ
du = dθ
dv = secθtanθ dθ
v = secθ
Int(θ secθtanθ) = θsecθ - Int(secθ dθ)
To integrate secθ you have to resort to a trick:
secθ (secθ + tanθ)/(secθ + tanθ) dθ
now the top is sec^2θ + secθtanθ
let u = secθ + tanθ and we have
1/u du
so, Int(secθ dθ) is ln|secθ + tanθ|
and we end up with
Int(θ secθtanθ) = θsecθ - ln|secInt(θ secθtanθ) = θsecθ - Int(secθ dθ) + tanInt(θ secθtanθ)
= θsecθ - Int(secθ dθ)|
Now, what does all that equal in x's?
θ = arcsec(x)
secθ = x
tanθ = √(x^2-1)
and you have your answer.
copy-paste error. That last complicated line should read:
Int(θ secθtanθ) = θsecθ - Int(secθ dθ)
= θsecθ - ln|secθ + tanθ|
To prove that the integral of arcsec(x) is equal to xarcsec(x) - ln|x + sqrt(x^2 - 1)|, we can start by finding the derivative of the expression on the right-hand side.
Let's consider the function f(x) = xarcsec(x) - ln|x + sqrt(x^2 - 1)|. We'll differentiate this function with respect to x and check if we get arcsec(x) as the result.
To differentiate f(x), we need to apply the chain rule and product rule.
First, let's differentiate the term xarcsec(x). Using the product rule:
f'(x) = x * d/dx(arcsec(x)) + arcsec(x) * d/dx(x)
To differentiate arcsec(x), we need to apply the chain rule. Let u = arcsec(x), and rewrite it as: x = sec(u), which gives us dx = sec(u) * tan(u) du.
Now, we can differentiate arcsec(x) using the chain rule:
d/dx(arcsec(x)) = d/du(arcsec(x)) * d/dx(sec(u)) = 1/(|x| * sqrt(x^2 - 1)) * sec(u) * tan(u)
Substituting the expressions for dx and d/dx(arcsec(x)) back into our original equation:
f'(x) = x * (1/(|x| * sqrt(x^2 - 1)) * sec(u) * tan(u)) + arcsec(x)
Note that we have sec(u) * tan(u) in the numerator and denominator, which simplifies to 1. Also, |x| is equal to x for x > 0 and -x for x < 0.
Simplifying further:
f'(x) = (x / (x * sqrt(x^2 - 1))) + arcsec(x)
f'(x) = (1 / sqrt(x^2 - 1)) + arcsec(x)
We see that f'(x) is equal to 1 / sqrt(x^2 - 1) + arcsec(x), which is equivalent to arcsec(x).
Therefore, we have shown that the derivative of f(x) is equal to arcsec(x). Since the derivative of f(x) is arcsec(x), we can conclude that the integral of arcsec(x) is xarcsec(x) - ln|x + sqrt(x^2 - 1)|.