# Geometrical interpretation of trigonometric antiderivative

I know about geometrical explanation of [defininte] integral as an area under the curve, and I wonder if there are some ideas, which may give similar insight in taking antiderivatives [indefinite integrals?].

Exactly, I have $\int \frac{\sin x \cos x}{\sin^4 x + \cos^4 x}$. It may be solved algebraically (by using $\sin 2x = 2 \sin x \cos x$ and $\sin^2 x + \cos^2 x = 1$ few times to get $\frac{\frac{1}{2} \sin 2x}{1 - \frac{1}{2} \sin^2 2x} = \frac{t}{2 - t^2}$ or something like this), which will give the result $-\frac{1}{2} \arctan(\cos 2x) + C$.

But how can it be solved geometrically, or by using geometric ideas? For example, may be it is possible to change the coordinate system to something like spherical coordinates or something more exotic, and then the answer will be evident? Or may be everything will be evident if I look on plot of some function?

I know this may have no practical value, but for some people geometrical insights are very interesting and motivating.

Any ideas and suggestions appeciated.

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Quoting you "For example, may be it is possible to change the coordinate system to something like spherical coordinates or something more exotic, and then the answer will be evident?" Thats what you have done when you let $t = \sin(2x)$. You have transformed your coordinate system into an "exotic" coordinate system by transforming $t = \sin(2x)$ –  user17762 Apr 21 '11 at 4:07
Well, that is true. But it is still not very clear (from geometrical point of view), what is the antiderivative of the expression with $t$ variable. –  Mixo123 Apr 21 '11 at 17:53