How do you integrate #int x / sqrt(-x^2+4x) dx# using trigonometric substitution?

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Answer 1 · 2016-03-23T20:29:07

#-sqrt(-x^2 +4x)+const.#

Consider that
#x^2 -4x+4=(x-2)^2#
Then
#x^2 -4x=-4+(x-2)^2#
And
#-x^2 +4x=4-(x-2)^2#

Rewriting the expression
#=int x/sqrt(4-(x-2)^2)dx#

Making
#(x-2)=2siny#
#dx=2cosy*dy#

The expression becomes

#=int (2siny*2cosy)/sqrt(4-(2siny)^2)dy#
#=int (4siny*cancel(cosy))/(2cancel(cosy))dy#
#=-2cos y+const.#

But
#siny=(x-2)/2#
#-> cos y =sqrt(1-sin^2 y)=sqrt(1-((x-2)/2)^2)=sqrt (-x^2+4x)/2#

So the result is
#=-sqrt(-x^2 +4x)+const.#