Question

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

Answer 1

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

Explanation:

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.`