Wednesday, 25 November 2015

calculus - Calculate intfrac1sqrt4x2dx



Calculate 14x2dx



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 = x2dudx=12




Then the integral became:



=11u2du=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



dx4x2=2 du4(2u)2=2 du4(1u2)=2 du21u2=du1u2



Why especially this substitution: Notice that



dx4x2=dx4(114x2)=dx21(x2)2




so you can see that it is quite nice to substitute u=x2; we get a function 11u2 and we already know the integral to this one.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...