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Is there a situation where it might be useful to integrate a function by taking the rectangle between the origin and the upper-right of the area, minus the rectangle between the origin and the lower-left of the area, minus the area to the left of the curve which could be calculated as the area under the inverse function. In other words:

$$ \int_a^b \! f(x) \, \textrm{d}x = b \cdot f(b) - a \cdot f(a) - \int_{f(a)}^{f(b)} \! f^{-1}(x) \, \textrm{d}x $$

Also, does this technique happen to have a name that I could lookup?

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This is just an example of how you can use it: $$\int_0^1 \arctan(x)dx=\frac{\pi}{4}-\int_0^{\pi/4}\tan x dx=\frac{\pi}{4}+\log\frac{1}{\sqrt2}.$$

Proof of the formula using integration by parts:

$$\int_a^b f(x)dx=xf(x)|_a^b-\int_a^bxf'(x)dx$$ now using the change of variable $y=f(x)$, with $dy=f'(x)$, one gets $$\int_a^b f(x)dx=xf(x)|_a^b-\int_{f(a)}^{f(b)}f^{-1}(y)dy.$$

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  • $\begingroup$ Interesting, I would've used integration by parts for $\int \! \arctan(x) \, \textrm{d}x$. Is there a relation between that technique and the technique I described? $\endgroup$ Commented Mar 20, 2016 at 18:35
  • $\begingroup$ @PrettyAntlers see my edit. $\endgroup$ Commented Mar 20, 2016 at 19:21

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