INTEGRATION 1/sqrt8x-x^2 dx?

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Answer 1 · 2018-05-28T12:48:27

#int dx/sqrt(8x-x^2) = arcsin((x-4)/4)+c#

Complete the square at the denominator:

#int dx/sqrt(8x-x^2) = int dx/sqrt(16-16+8x-x^2) #

#int dx/sqrt(8x-x^2) = int dx/sqrt(16-(x-4)^2)#

#int dx/sqrt(8x-x^2) = 1/4 int dx/sqrt(1-((x-4)/4)^2)#

Substitute:

#(x-4) = 4 sint#

#dx = 4costdt#

with #t in (-pi/2,pi/2)# to have:

#int dx/sqrt(8x-x^2) = 1/cancel 4 int (cancel4costdt)/sqrt(1-sin^2t)#

and considering that in the interval where #t# ranges, #cost >0#:

#sqrt(1-sin^2t) = sqrt(cos^2t) = cost#

so:

#int dx/sqrt(8x-x^2) = int dt = t+c#

undoing the substitution:

#int dx/sqrt(8x-x^2) = arcsin((x-4)/4)+c#