How do you integrate int x / sqrt(-x^2+4x) dx∫x√−x2+4xdx using trigonometric substitution?

1 Answer
Rui D.
Mar 23, 2016

-sqrt(-x^2 +4x)+const.−√−x2+4x+const.

Explanation:

Consider that
x^2 -4x+4=(x-2)^2x2−4x+4=(x−2)2
Then
x^2 -4x=-4+(x-2)^2x2−4x=−4+(x−2)2
And
-x^2 +4x=4-(x-2)^2−x2+4x=4−(x−2)2

Rewriting the expression
=int x/sqrt(4-(x-2)^2)dx=∫x√4−(x−2)2dx

Making
(x-2)=2siny(x−2)=2siny
dx=2cosy*dydx=2cosy⋅dy

The expression becomes

=int (2siny*2cosy)/sqrt(4-(2siny)^2)dy=∫2siny⋅2cosy√4−(2siny)2dy
=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.

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