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Real quick: If I have the function $$\int { \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } dx$$

I can easily substitute by setting $x$ equal to $a\sin \theta$.

But why actually is that? If I draw a right triangle it can also be $x = a\cos \theta$, depending on where you choose the sides and the angle $\theta$ to be.

If it has nothing to do with it then I ask myself why people derive trig substitution by drawing a right triangle...

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  • $\begingroup$ You may solve the integral with substitution of $x=acos(\theta)$, what is your problem? $\endgroup$ – Majid Jun 28 '16 at 19:24
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You can use either. Remember, though, that in the integration when you use $x = a \sin \theta$ you also need to substitute $dx = (\cos \theta) d \theta$. If you were to use $x = a \cos \theta$ then you would instead need to substitute $dx = -(\sin \theta) d \theta$. You will get the same final answer either way. The reason you probably see the former substitution of $x = a \sin \theta$ more commonly is that you avoid the negative sign in the substitution and the work might be slightly cleaner looking that way.

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