# How to Integrate Sqrt(Tan(x)) | Hard Antiderivatives

In general, the square root of a trig function is not integrable, for example, sqrt(sin(x)) and sqrt(cos(x)) are functions whose antiderivatves cannot be expressed in closed form. However, there are some notable exceptions, such as sqrt(1 ± sin(x)), sqrt(1 ± cos(x)), and sqrt(tan(x)). The square root of the tangent of x is an integrable function, though it does take some work to figure out its antiderivative. One way to find the antiderivative of sqrt(tan(x)) is with substitution, which yields a quartic rational function.

Since rational functions are always integrable, and their antiderivatives involve logarithms and inverse tangent functions, you will end up with an antiderivative involving Ln() and arctan().

## Step 1: Substitution

The most straight-forward (but nonetheless time-consuming) way to find the antiderivative of sqrt(tan(x)) is to use the substitution x = arctan(z^2) and dx = 2z/(z^4 + 1) dz. This gives the integral equality

∫ sqrt(tan(x)) dx = ∫ (2z^2)/(z^4 + 1) dz

This transforms a radical and trigonometric function into a rational function. From the tutorial on how to integrate the function f(x) = 1/(x^4 + 1), we know that the quartic polynomial z^4 + 1 factors into (z^2 + sqrt(2)z + 1) * (z^2 - sqrt(2)z + 1), which means we can use the technique of partial fractions to decompose (2z^2)/(z^4 + 1) into a sum of simpler rational functions

(2z^2)/(z^4 + 1) =

(Az + B)/(z^2 + sqrt(2)z + 1) + (Cz + D)/(z^2 - sqrt(2)z + 1).

## Step 2: Partial Fractions

To find the constants A, B, C, and D that produce the correct decomposition, we need to solve the equation

(Az + B)(z^2 - sqrt(2)z + 1) + (Cz + D)(z^2 + sqrt(2) + 1) = 2z^2

for A, B, C, and D. This produces a linear system of four equations in four variables:

A + C = 0

-sqrt(2)A + B + sqrt(2)C + D = 2

A - sqrt(2)B + C + sqrt(2)D = 0

B + D = 0

The solution to this system is A = -1/sqrt(2), B = 0, C = 1/sqrt(2), and D = 0.

## Step 3: Integration

Given that (2z^2)/(z^4 + 1) is equivalent to the sum

(-z/sqrt(2))/(z^2 + sqrt(2)z + 1) + (z/sqrt(2))/(z^2 - sqrt(2)z + 1)

we can put this rational function into an even more integration friendly form:

(-z/sqrt(2) - 1/2)/(z^2 + sqrt(2)z + 1) + (z/sqrt(2) - 1/2)/(z^2 - sqrt(2)z + 1)

+ (1/2)/(z^2 + sqrt(2)z + 1) + (1/2)/(z^2 - sqrt(2)z + 1)

The first two summands are recognizable as derivatives of logarithmic functions, while the last two are derivatives of inverse tangent functions. [See How to Integrate 1/(ax^2 + bx + c)] Integrating this expression piece by piece gives us

∫ (-z/sqrt(2) - 1/2)/(z^2 + sqrt(2)z + 1) dz

= (-sqrt(2)/4)*Ln(z^2 + sqrt(2)z + 1) + C

∫ (z/sqrt(2) - 1/2)/(z^2 - sqrt(2)z + 1) dz

= (sqrt(2)/4)*Ln(z^2 - sqrt(2)z + 1) + C

∫ (1/2)/(z^2 + sqrt(2)z + 1) dz

= (sqrt(2)/2)*arctan(sqrt(2)z + 1) + C

∫ (1/2)/(z^2 - sqrt(2)z + 1) dz

= (sqrt(2)/2)*arctan(sqrt(2)z - 1) + C

## Step 4: Back Substitution

Since z = sqrt(tan(x)), we need to back-substitute in the z-variable integrals to find the antiderivative of sqrt(tan(x)). Doing so gives us the final expression for the integral of sqrt(tan(x)):

∫ sqrt(tan(x)) dx =

(-sqrt(2)/4)*Ln[tan(x) + sqrt(2*tan(x)) + 1] + (sqrt(2)/4)*Ln[tan(x) - sqrt(2*tan(x)) + 1] + (sqrt(2)/2)*arctan[sqrt(2*tan(x)) + 1] + (sqrt(2)/2)*arctan[sqrt(2*tan(x)) - 1] + C

## Example Integration

Find the area under the curve y = sqrt(tan(x)) from x = 0 to x = π/2.

The endpoint x = π/2 is a vertical asymptote of the function (as is x = -π/2, x = 3π/2, etc.) so the integral may not converge, that is, the area could be infinite. The only way to see if the area is finite or infinite is the plug the limits of integration into the antiderivative and subtract. For this problem, we will use an equivalent form of the antiderivative:

∫ sqrt(tan(x)) dx =

(sqrt(2)/2)*Ln[sin(x) + cos(x) - sqrt(sin(2x))] + (sqrt(2)/2)*arctan[sqrt(2*tan(x)) + 1] +

(sqrt(2)/2)*arctan[sqrt(2*tan(x)) - 1] + C

The logarithmic function in this form is simplified by using properties of logarithms and trig identities. Plugging x = 0 and x = π/2 into the antiderivative and subtracting gives you

{ (sqrt(2)/2)*Ln[sin(π/2) + cos(π/2) - sqrt(sin(π))] + (sqrt(2)/2)*arctan[sqrt(2*tan(π/2)) + 1] +(sqrt(2)/2)*arctan[sqrt(2*tan(π/2)) - 1] } -

{ (sqrt(2)/2)*Ln[sin(0) + cos(0) - sqrt(sin(0))] + (sqrt(2)/2)*arctan[sqrt(2*tan(0)) + 1] +(sqrt(2)/2)*arctan[sqrt(2*tan(0)) - 1] } =

{ (sqrt(2)/2)*Ln[1] + (sqrt(2)/2)*arctan[∞] + (sqrt(2)/2)*arctan[∞] } -

{ (sqrt(2)/2)*Ln[1] + (sqrt(2)/2)*arctan[1] + (sqrt(2)/2)*arctan[-1] } =

**π/sqrt(2)**.

As it turns out, this improper integral does converge to a finite area.

## Related Integrals

This substitution method can also be used to resolve similar trigonometric integrands involving square roots. Here are the integrals of sqrt(cot(x)) = 1/sqrt(tan(x)) as well as cos(x)*sqrt(sin(2x)) and sin(x)*sqrt(sin(2x)). As you can see from their antiderivaitves, these functions are all very closely related.

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