# Evaluate the indefinite integral. integrate of sqrt(20x-x^2)dx

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### 1 Answer

You should complete the square such that:

`sqrt(-(100 -20x + x^2) + 100) = sqrt(100 - (x - 10)^2)`

You need to come up with the following substitution such that:

`x - 10 = t => dx = dt`

`int sqrt(20 - x^2)dx = int sqrt(100 - t^2)dt `

You should remember the following formula such that:

`int sqrt(a^2-x^2)dx = (1/2)(xsqrt(a^2-x^2) + a^2arcsin (x/a))`

Reasoning by analogy yields:

`int sqrt(100 - t^2)dt = (1/2)(t*sqrt(100-t^2) + 100arcsin (t/10)) + c`

Substituting back `x - 10` for t yields:

`int sqrt(20 - x^2)dx = (1/2)((x-10)*sqrt(100-(x-10)^2) + 100arcsin ((x-10)/10)) + c`

**Hence, evaluating the integral of irrational function `sqrt(20 - x^2)` yields `int sqrt(20 - x^2)dx = (1/2)((x-10)*sqrt(100-(x-10)^2) + 100arcsin ((x-10)/10)) + c.` **