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How to Integrate Sqrt(x^2 - 1) and Sqrt(1 - x^2)

Updated on April 15, 2015
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TR Smith is a teacher and creator who uses mathematics in her line of work every day.

Integrals of sqrt(x^2 - 1) and sqrt(1 - x^2)
Integrals of sqrt(x^2 - 1) and sqrt(1 - x^2)

Typographically, the functions f(x) = sqrt(x^2 - 1) and g(x) = sqrt(1 - x^2) look very similar, but mathematically and graphically there are some key differences. The function f(x) = sqrt(x^2 - 1) is defined only for values of x that are less than or equal to -1 or greater than or equal to 1. In contrast, g(x) = sqrt(x^2 - 1) is only defined for values of x between -1 and 1 (including -1 and 1).

The function sqrt(x^2 - 1) is the upper half of a hyperbola, a conic section. The function sqrt(1 - x^2) is the upper half of the unit circle, another conic section. Both curves are shown in the graph below.

Similar techniques involving trig substitution are used to integrate the functions sqrt(x^2 - 1) and sqrt(1 - x^2).


Graphs of y = sqrt(1 - x^2) in red and y = sqrt(x^2 - 1) in blue.
Graphs of y = sqrt(1 - x^2) in red and y = sqrt(x^2 - 1) in blue.

Integrating Sqrt(1 - x^2), the Unit Circle

The first step in finding the antiderivative of the function y = sqrt(1 - x^2) is to apply a trigonometric substitution. We let

x = sin(u)
dx = cos(u) du

This transforms our integral from a radical function to a simple trig function:

What is sin(2*arcsin(x))?

To convert sin(2*arcsin(x)) to an algebraic function, we recall the double angle formula for sine, sin(2θ) = 2sin(θ)cos(θ). This gives us:

sin(2*arcsin(x))
= 2*sin(arcsin(x))*cos(arcsin(x))
= 2*x*sqrt[ 1 - sin(arcsin(x)^2 ]
= 2x*sqrt[1 - x^2]

The penultimate line comes from the identity cos(θ) = sqrt[1 - sin(θ)^2].

∫ sqrt(1 - x^2) dx
= ∫ sqrt(1 - sin(u)^2) cos(u) du
= ∫ sqrt(cos(u)^2) cos(u) du
= ∫ cos(u)^2 du

How to find integral of cosine squared was explained in a previous article. It involves using the trig identity cos(u)^2 = [1 + cos(2u)]/2. Sparing the details, we can just go straight to the antiderivative of cosine squared.

∫ cos(u)^2 du
= u/2 + sin(2u)/4 + c

Making the reverse substitution with u = arcsin(x) gives us

arcsin(x)/2 + sin(2*arcsin(x))/4 + c

We can use another trig identity to simplify this expression further: sin(2*arcsin(x)) = 2x*sqrt(1 - x^2). This produces the standard form of the integral of sqrt(1 - x^2).

∫ sqrt(1 - x^2) = arcsin(x)/2 + (x/2)*sqrt(1 - x^2) + c


Integrating Sqrt(x^2 - 1), a Hyperbola

Integrating sqrt(x^2 - 1) is similar to integrating sqrt(1 - x^2) in that the first step is a trig substitution. For this integral, we need to use

x = sec(u)
dx = sec(u)tan(u) du

This turns our original radical function into a trig function:

What is tan(arcsec(x))?

To simplify tan(arcsec(x)), we use the identities and tan(θ) = sqrt[sec(θ)^2 - 1] and sec(arcsec(w)) = w. This gives us

tan(arcsec(x))
= sqrt[sec(arcsec(x))^2 - 1]
= sqrt[w^2 - 1]

∫ sqrt(x^2 - 1) dx
= ∫ sqrt(sec(u)^2 - 1) sec(u)tan(u) du
= ∫ sqrt(tan(u)^2) sec(u)tan(u) du
= ∫ sec(u)tan(u)^2 du
= ∫ sec(u) * (sec(u)^2 - 1) du
= ∫ sec(u)^3 du - ∫ sec(u) du

The integral of sec(u)^3 is worked out here and also in the article for finding the antiderivative of sqrt(x^2 + a^2), while the integral of sec(u) is worked out in another tutorial. If we combine the results of these previous articles, we get

∫ sec(u)^3 du - ∫ sec(u) du
= 0.5*tan(u)sec(u) + 0.5*Ln[sec(u) + tan(u)] - Ln[sec(x) + tan(x)] + c
= 0.5*tan(u)sec(u) - 0.5*Ln[sec(u) + tan(u)] + c

Now we just have to make the reverse substitution u = arcsec(x):

0.5*tan(arcsec(x))*x - 0.5*Ln[x + tan(arcsec(x))] + c

Using the trig identity tan(arcsec(x)) = sqrt(x^2 - 1), we get

∫ sqrt(x^2 - 1) dx = (x/2)*sqrt(x^2 - 1) - 0.5*Ln[x + sqrt(x^2 - 1)] + c


Example 1

Find the area under the unit circle between x = 0.2 and x = 0.7. The region in question is shaded in pink in the graph below.

Since the antiderivative of the unit circle curve is

arcsin(x)/2 + (x/2)*sqrt(1 - x^2) + c

we simply need to plug the endpoints x = 0.7 and x = 0.2 into the formula and subtract in order to find the area of the curve. Remember that arcsin is in radians, not degrees.

[arcsin(0.7)/2 + (0.7/2)*sqrt(1 - 0.49)] - [arcsin(0.2)/2 + (0.2/2)*sqrt(1 - 0.04)]

= 0.43899


Example 2

Solve the separable differential eqaution

dy/dx = sqrt[ (x^2 - 1)/y ].

Since this equation can be rewritten as

dy/dx = sqrt(x^2 - 1) / sqrt(y),

separating the variables gives us

sqrt(y) dy = sqrt(x^2 - 1) dx

Fortunately we can integrate both sides of this equation because we now know the antiderivative of the expression on the right-hand side.

∫ sqrt(y) dy = ∫ sqrt(x^2 - 1) dx

(2/3)y^(3/2) = (x/2)*sqrt(x^2 - 1) - 0.5*Ln[x + sqrt(x^2 - 1)] + c

y^(3/2) = (3x/4)*sqrt(x^2 - 1) - 0.75*Ln[x + sqrt(x^2 - 1)] + 1.5*c

y = { (3x/4)*sqrt(x^2 - 1) - 0.75*Ln[x + sqrt(x^2 - 1)] + 1.5*c }^(2/3)

y = { (3x/4)*sqrt(x^2 - 1) - 0.75*Ln[x + sqrt(x^2 - 1)] + C }^(2/3)

The bolded equation above is the solution to this differential equation for any arbitrary constant C.


Example 3

Is the area between the curves y = x and y = sqrt(x^2 - 1) from x = 1 to x = ∞ finite or unbounded? To answer this question we need to evaluate the integral

∫ x - sqrt(x^2 - 1) dx

from x = 1 to x = infinity. The antiderivative of the integrand is

(1/2){ x^2 - x*sqrt(x^2 - 1) + Ln[x + sqrt(x^2 - 1)] }

Evaluated at x = 1, the value of this expression is 1/2. The limit of this expression as x goes to infinity is infinity. Therefore, the improper integral does not converge and the area between the curves is infinite.

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    • ian 2 years ago

      what is cos(2*arccos(x)) and cos(0.5*arccos(x))?

    • calculus-geometry profile image
      Author

      TR Smith 2 years ago from Germany

      cos(2*arccos(x)) = 2x^2 - 1, and cos(arccos(x)/2) = sqrt[(x+1)/2]. These functions are inverses of each other.

    • dean 3 months ago

      integrate (1+x^2)^1/2/(1-x^2)

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