Challenge integral (2 Viewers)

unofficiallyred12 · Oct 11, 2023

but ye my teacher said such questions aren't likely to come up in hsc cos could be controversial

Luukas.2 · Oct 11, 2023

The problem with the approach from @blob063540 arises from the initial transformation of the integrand

$\sqrt{\tan{x}} + \frac{1}{\sqrt{\tan{x}}} = \sqrt{\frac{\sin{x}}{\cos{x}}} + \sqrt{\frac{\cos{x}}{\sin{x}}} = \frac{\sqrt{\sin^2{x}} + \sqrt{\cos^2{x}}}{\sqrt{\sin{x}\cos{x}}} = \frac{\sin{x} + \cos{x}}{\sqrt{\sin{x}\cos{x}}}$

It is true when $x$ lies in quadrant 1, but not when it is in quadrant 3 (because sine and cosine are negative whereas the tangent function is positive). The integrand itself is not defined in quadrants 2 or 4.

It arises because $\sqrt{\sin^2{x}} = \sin{x}$ only when $\sin{x} > 0$ . MX2 students should be aware that

$\sqrt{x^2} = |x| \qquad \implies \qquad \sqrt{\sin^2{x}} = |\sin{x}|$

A closed form that does work for all $x$ in the domain of the integrand is then

$\int \sqrt{\tan{x}} + \sqrt{\cot{x}}\ dx = \sqrt{2}\sin^{-1}{\big(\left|\sin{x}\right| - \left|\cos{x}\right|\big)} + C$

Interestingly, given the discussion above, this integrated form is defined for all real $x$ , though it is not differentiable for any integer multiple of pi on 2. The function's behaviour in the second and fourth quadrants does not have meaning in terms of the original integral.

The form I derived earlier, that

$\int \sqrt{\tan{x}} + \sqrt{\cot{x}}\ dx = \sqrt{2} \left[\tan^{-1}{\left(\sqrt{2\tan{x}} + 1\right)} + \tan^{-1}{\left(\sqrt{2\tan{x}} - 1\right)}\right] + C$

is defined only when $\tan{x} \ge 0$ , as are the form from @member 6003's approach, that

$\begin{align*} \int \sqrt{\tan{x}} + \sqrt{\cot{x}}\ dx &= \sqrt{2}\tan^{-1}{\left[\frac{1}{\sqrt{2}}\left(\sqrt{\tan{x}} - \cfrac{1}{\sqrt{\tan{x}}}\right)\right]} + C \\ &= \sqrt{2}\tan^{-1}{\left(\frac{\tan{x} - 1}{\sqrt{2\tan{x}}}\right)} + C \end{align*}$

The final form has used $\sqrt{\tan^2{x}} = \tan{x}$ which is accurate in this case as the presence of $\sqrt{2\tan{x}}$ in the denominator ensures that the domain is restricted to when $\tan{x} > 0$ . Thus, formally, the reasoning is:

$\begin{align*} \sqrt{2}\tan^{-1}{\left[\frac{1}{\sqrt{2}}\left(\sqrt{\tan{x}} - \cfrac{1}{\sqrt{\tan{x}}}\right)\right]} &= \sqrt{2}\tan^{-1}{\left(\frac{\sqrt{\tan^2{x}} - 1}{\sqrt{2\tan{x}}}\right)} \\ &= \sqrt{2}\tan^{-1}{\left(\frac{\left|\tan{x}\right| - 1}{\sqrt{2\tan{x}}}\right)} \\ &= \sqrt{2}\tan^{-1}{\left(\frac{\tan{x} - 1}{\sqrt{2\tan{x}}}\right)} \qquad \text{as the result is only defined when $\tan{x} > 0$} \end{align*}$

Any of the forms presented here is valid. There is no problem with member 6003's approach or working, in my view.

member 6003 · Oct 12, 2023

Luukas.2 said:
The problem with the approach from @blob063540 arises from the initial transformation of the integrand

$\sqrt{\tan{x}} + \frac{1}{\sqrt{\tan{x}}} = \sqrt{\frac{\sin{x}}{\cos{x}}} + \sqrt{\frac{\cos{x}}{\sin{x}}} = \frac{\sqrt{\sin^2{x}} + \sqrt{\cos^2{x}}}{\sqrt{\sin{x}\cos{x}}} = \frac{\sin{x} + \cos{x}}{\sqrt{\sin{x}\cos{x}}}$

It is true when $x$ lies in quadrant 1, but not when it is in quadrant 3 (because sine and cosine are negative whereas the tangent function is positive). The integrand itself is not defined in quadrants 2 or 4.

It arises because $\sqrt{\sin^2{x}} = \sin{x}$ only when $\sin{x} > 0$ . MX2 students should be aware that

$\sqrt{x^2} = |x| \qquad \implies \qquad \sqrt{\sin^2{x}} = |\sin{x}|$

A closed form that does work for all $x$ in the domain of the integrand is then

$\int \sqrt{\tan{x}} + \sqrt{\cot{x}}\ dx = \sqrt{2}\sin^{-1}{\big(\left|\sin{x}\right| - \left|\cos{x}\right|\big)} + C$

Interestingly, given the discussion above, this integrated form is defined for all real $x$ , though it is not differentiable for any integer multiple of pi on 2. The function's behaviour in the second and fourth quadrants does not have meaning in terms of the original integral.

The form I derived earlier, that

$\int \sqrt{\tan{x}} + \sqrt{\cot{x}}\ dx = \sqrt{2} \left[\tan^{-1}{\left(\sqrt{2\tan{x}} + 1\right)} + \tan^{-1}{\left(\sqrt{2\tan{x}} - 1\right)}\right] + C$

is defined only when $\tan{x} \ge 0$ , as are the form from @member 6003's approach, that

$\begin{align*} \int \sqrt{\tan{x}} + \sqrt{\cot{x}}\ dx &= \sqrt{2}\tan^{-1}{\left[\frac{1}{\sqrt{2}}\left(\sqrt{\tan{x}} - \cfrac{1}{\sqrt{\tan{x}}}\right)\right]} + C \\ &= \sqrt{2}\tan^{-1}{\left(\frac{\tan{x} - 1}{\sqrt{2\tan{x}}}\right)} + C \end{align*}$

The final form has used $\sqrt{\tan^2{x}} = \tan{x}$ which is accurate in this case as the presence of $\sqrt{2\tan{x}}$ in the denominator ensures that the domain is restricted to when $\tan{x} > 0$ . Thus, formally, the reasoning is:

$\begin{align*} \sqrt{2}\tan^{-1}{\left[\frac{1}{\sqrt{2}}\left(\sqrt{\tan{x}} - \cfrac{1}{\sqrt{\tan{x}}}\right)\right]} &= \sqrt{2}\tan^{-1}{\left(\frac{\sqrt{\tan^2{x}} - 1}{\sqrt{2\tan{x}}}\right)} \\ &= \sqrt{2}\tan^{-1}{\left(\frac{\left|\tan{x}\right| - 1}{\sqrt{2\tan{x}}}\right)} \\ &= \sqrt{2}\tan^{-1}{\left(\frac{\tan{x} - 1}{\sqrt{2\tan{x}}}\right)} \qquad \text{as the result is only defined when $\tan{x} > 0$} \end{align*}$

Any of the forms presented here is valid. There is no problem with member 6003's approach or working, in my view.

I don't know how you got

$\int \sqrt{\tan x}+\sqrt{\cot x} d x=\sqrt{2} \sin ^{-1}(|\sin x|-|\cos x|)+C$

it looks wrong in desmos.
I'll list what did to resolve the problem with blob's answer

$\int \sqrt{\tan x} +\sqrt{\cot x} dx$

$= \int \sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}} dx$

$= \int \frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}} dx$

This step is actually valid since the integrand implies sinx, cosx have the same sign, if they're both negative then it will cancel under the square root so you can use the property $\sqrt{ab}=\sqrt{a}\sqrt{b}$ which doesn't work for complex numbers.

$= \int \frac{\sqrt{\sin^2 x}+\sqrt{\cos^2 x}}{\sqrt{\cos x \sin x}} dx$

$= \int \frac{|\sin x|+|\cos x|}{\sqrt{\cos x \sin x}} dx$

since we know that sinx and cosx are either both positive or negative:
case 1 is for $\cos x , \sin x \geq 0$
you get

$\int \sqrt{\tan x}+\sqrt{\cot x} d x=\sqrt{2}\left[\sin ^{-1}(\sin x-\cos x)\right]+C$

case 2 is for $\cos x , \sin x < 0$
you get

$\int \sqrt{\tan x}+\sqrt{\cot x} d x=\sqrt{2}\left[\sin ^{-1}(\cos x-\sin x)\right]+C$

interestingly enough you don't have to apply both restrictions only one of them because of the restrictions of inverse sine. You can try to combine the two cases using the sign function but then you get zeroes where there shouldn't be in the case that sinx or cosx=0

Luukas.2 · Oct 12, 2023

member 6003 said:
I don't know how you got

$\int \sqrt{\tan x}+\sqrt{\cot x} d x=\sqrt{2} \sin ^{-1}(|\sin x|-|\cos x|)+C$
it looks wrong in desmos.
I'll list what did to resolve the problem with blob's answer

$\int \sqrt{\tan x} +\sqrt{\cot x} dx$

$= \int \sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}} dx$

$= \int \frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}} dx$

This step is actually valid since the integrand implies sinx, cosx have the same sign, if they're both negative then it will cancel under the square root so you can use the property $\sqrt{ab}=\sqrt{a}\sqrt{b}$ which doesn't work for complex numbers.

$= \int \frac{\sqrt{\sin^2 x}+\sqrt{\cos^2 x}}{\sqrt{\cos x \sin x}} dx$

$= \int \frac{|\sin x|+|\cos x|}{\sqrt{\cos x \sin x}} dx$

since we know that sinx and cosx are either both positive or negative:
case 1 is for $\cos x , \sin x \geq 0$
you get

$\int \sqrt{\tan x}+\sqrt{\cot x} d x=\sqrt{2}\left[\sin ^{-1}(\sin x-\cos x)\right]+C$
case 2 is for $\cos x , \sin x < 0$
you get

$\int \sqrt{\tan x}+\sqrt{\cot x} d x=\sqrt{2}\left[\sin ^{-1}(\cos x-\sin x)\right]+C$
interestingly enough you don't have to apply both restrictions only one of them because of the restrictions of inverse sine. You can try to combine the two cases using the sign function but then you get zeroes where there shouldn't be in the case that sinx or cosx=0

Yes, another approach is to express the integral as piece-wise defined:

$\int \sqrt{\tan{x}} + \sqrt{\cot{x}}\ dx = \sqrt{2}\left[\sin^{-1}{(\sin{x} - \cos{x})}\right] + C \qquad \text{provided $x$ is an angle in the first quadrant}$

$\int \sqrt{\tan{x}} + \sqrt{\cot{x}}\ dx = \sqrt{2}\left[\sin^{-1}{(\cos{x} - \sin{x})}\right] + C \qquad \text{provided $x$ is an angle in the third quadrant}$

If you look using DESMOS carefully, you will find that this piece-wise definition exactly matches your inverse tan solutions.

The solution I presented with the absolute values also exactly matches all these solutions for all angles in quadrants 1 and 3 (including the boundaries). It then gives a reflection in $x = \pi / 2$ "solution" for quadrants 2 and 4, but these don't matter as the integrand is not defined in these quadrants.

Luukas.2 · Oct 12, 2023

By the way, the @blob063540 approach is more rigorously re-written as:

$\begin{align*} \int \sqrt{\tan{x}} + \sqrt{\cot{x}}\ dx &= \int \sqrt{\tan{x}} + \frac{1}{\sqrt{\tan{x}}}\ dx \\ &= \int \frac{\left(\sqrt{\tan{x}}\right)^2 + 1}{\sqrt{\tan{x}}}\ dx \\ &= \int \frac{\left|\tan{x}\right| + 1}{\sqrt{\tan{x}}} \times \frac{\left|\cos{x}\right|}{\left|\cos{x}\right|}\ dx \\ &= \int \frac{\left|\sin{x}\right| + \left|\cos{x}\right|}{\sqrt{\tan{x}} \times \left|\cos{x}\right|}\ dx \\ &= \int \frac{\left|\sin{x}\right| + \left|\cos{x}\right|}{\sqrt{\tan{x} \times \cos^2{x}}}\ dx \\ &= \int \frac{\left|\sin{x}\right| + \left|\cos{x}\right|}{\sqrt{\sin{x}\cos{x}}}\ dx \end{align*}$

This avoids the problem of needing to deal with a square root of a negative (as MX2 does not actually cover complex number calculus), because in the domain of the integrand, $\sin{x}\cos{x}$ can't be negative.

carrotsss · Oct 12, 2023

scaryshark09 · Oct 12, 2023

Challenge integral (2 Viewers)

unofficiallyred12

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Luukas.2

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member 6003

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Luukas.2

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Luukas.2

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carrotsss

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scaryshark09

∞∆ who let 'em cook dis long ∆∞

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