The antiderivative of 1/(x^4+1): Part 9

In the course of evaluating the antiderivative

\displaystyle \int \frac{1}{x^4 + 1} dx,

I have stumbled across a very curious trigonometric identity:

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi if x < x_2,

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) if x_2 < x < x_1,

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi if x> x_1,

where x_1 and x_2 are the unique values so that

\tan^{-1} ( x_1\sqrt{2} - 1 ) + \tan^{-1}( x_1 \sqrt{2} + 1) = \displaystyle \frac{\pi}{2},

\tan^{-1} ( x_2\sqrt{2} - 1 ) + \tan^{-1}( x_2 \sqrt{2} + 1) = -\displaystyle \frac{\pi}{2}.

I will now show that x_1 = 1 and x_2 = -1. Indeed, it’s apparent that these have to be the two transition points because these are the points where \displaystyle \frac{x \sqrt{2}}{1 - x^2} is undefined. However, it would be more convincing to show this directly.

To show that x_1 = 1, I need to show that

\tan^{-1} (\sqrt{2} - 1 ) + \tan^{-1}( \sqrt{2} + 1) = \displaystyle \frac{\pi}{2}.

I could do this with a calculator…

arctangent…but that would be cheating.

Instead, let \alpha = \tan^{-1} (\sqrt{2} - 1 ) and \beta = \tan^{-1} (\sqrt{2} + 1 ), so that

\tan \alpha = \sqrt{2} - 1,

\tan \beta = \sqrt{2} + 1.

Indeed, by SOHCAHTOA, the angles \alpha and \beta can be represented in the figure below:

arctangenttriangle2The two small right triangles make one large triangle, and I will show that the large triangle is also a right triangle. To do this, let’s find the lengths of the three sides of the large triangle. The length of the longest side is clearly \sqrt{2} - 1 + \sqrt{2} + 1 = 2\sqrt{2}. I will use the Pythagorean theorem to find the lengths of the other two sides. For the small right triangle containing \alpha, the missing side is

\sqrt{ \left(\sqrt{2} - 1 \right)^2 + 1^2} = \sqrt{2 - 2\sqrt{2} + 1 + 1} = \sqrt{4-2\sqrt{2}}

Next, for the small right triangle containing \beta, the missing side is

\sqrt{ \left(\sqrt{2} + 1 \right)^2 + 1^2} = \sqrt{2 + 2\sqrt{2} + 1 + 1} = \sqrt{4+2\sqrt{2}}

So let me redraw the figure, eliminating the altitude from the previous figure:

arctangenttriangle3

Notice that the condition of the Pythagorean theorem is satisfied, since

\left( \sqrt{4-2\sqrt{2}} \right)^2 + \left( \sqrt{4+2\sqrt{2}} \right)^2 = 4 - 2\sqrt{2} + 4 + 2 \sqrt{2} = 8,

or

\left( \sqrt{4-2\sqrt{2}} \right)^2 + \left( \sqrt{4+2\sqrt{2}} \right)^2 = \left( 2\sqrt{2} \right)^2.

Therefore, by the converse of the Pythagorean theorem, the above figure must be a right triangle (albeit a right triangle with sides of unusual length), and so \alpha + \beta = \pi/2. In other words, x_1 = 1, as required.

To show that x_2 = -1, I will show that the function f(x) = \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) is an odd function using the fact that \tan^{-1} x is also an odd function:

f(-x) = \tan^{-1} ( -x\sqrt{2} - 1 ) + \tan^{-1}( -x \sqrt{2} + 1)

= \tan^{-1} ( -[x\sqrt{2} + 1] ) + \tan^{-1}( -[x \sqrt{2} - 1])

= -\tan^{-1} ( x\sqrt{2} + 1 ) - \tan^{-1}( x \sqrt{2} - 1)

= - \left[ \tan^{-1} ( x\sqrt{2} + 1 ) + \tan^{-1}( x \sqrt{2} - 1) \right]

= -f(x).

Therefore, f(-1) = -f(1) = -\displaystyle \frac{\pi}{2}, and so x_2 = -1.

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1 Comment

  1. The antiderivative of 1/(x^4+1): Index | Mean Green Math

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