Calculate ∫1√4−x2dx
Suppose that I only know regular substitution, not trig.
I tried to get help from an integral calculator, and what they did was:
Let u = x2→dudx=12
Then the integral became:
=∫1√1−u2du=arcsin(u)=arcsin(x2)
And I'm not sure how they accomplished this, where did the 4 go? I understand the arcsin part but not sure how they got rid of the 4? Also how did they know to substitute x2? It doesn't seem very obvious to me.
Answer
∫dx√4−x2=∫2 du√4−(2u)2=∫2 du√4(1−u2)=∫2 du2√1−u2=∫du√1−u2
Why especially this substitution: Notice that
∫dx√4−x2=∫dx√4(1−14x2)=∫dx2√1−(x2)2
so you can see that it is quite nice to substitute u=x2; we get a function 1√1−u2 and we already know the integral to this one.
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