When you see an integral in the form of shape of
\(\int a^2-x^2 dx\) or \(\int \sqrt{a^2-x^2} dx\),
you can solve it with trigonometric substitution. This is because, if \(a\) was the hypotenuse of the triangle, and \(x\) was the opposite side, then because of Pythagoras theorem the square of the adjacent side would be
\(a^2-x^2\), so the adjacent side is \(\sqrt{a^2-x^2}\).
For instance, \(\int \frac{1}{\sqrt{a^2-x^2}} dx\) can be solved easily if you make \(x = a sin \theta\). Then \(dx = a cos \theta d\theta\).
The angle of \(\theta \) is \(\theta = arcsin \frac{x}{a}\) since \(sin \theta = \frac{x}{a}\) in the triangle.
So you substitute the terms of the integral to obtain that
\(\int \frac{1}{\sqrt{a^2-x^2}} dx = \int \frac{acos \theta d \theta}{\sqrt{a^2-a^2sin^2\theta}} = \int \frac{acos \theta d \theta}{\sqrt{a^2(1-sin^2 \theta)}} =\)
\(= \int \frac{acos \theta d \theta}{\sqrt{a^2cos^2 \theta}} = \int \frac{acos \theta d \theta}{acos \theta} = \)
\(=\int d \theta = \theta + C = arcsin \frac{x}{a}+C \)
Which is where the integral identity
\(\int \frac{du}{\sqrt{a^2-u^2}} = arcsin \frac{u}{a}+C\)
comes from. Likewise you can use \(x=(a)tan \theta, \theta=arctan \frac{x}{a}\) for \(a^2+x^2\) and \(x=(a)sec \theta, \theta = arcsec \frac{x}{a}\) for \(x^2-a^2\), to get to the other two identities.
Izuru Yakumo (八雲 出流)
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