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

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### 2 Answers

You should use trigonometric substitution to solve the integral, but you need to convert the radicand into a difference of two squares by completing the square `20 x - x^2` such that:

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

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

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

`-x^2 + 20x - 100 + 100 = 100(1 - (x-10)^2/100)`

You should use the following substitution such that:

`(x - 10)/10 = sin t => dx = 10cos t dt`

Changing the variable yields:

`int sqrt(20x-x^2)dx = int (sqrt(100(1 - sin^2 t)))10cos t dt`

`int (sqrt(100(1 - sin^2 t)))10cos t dt = 100 int sqrt(1 - sin^2 t)*cos t dt`

You need to use the fundamental formula of trigonometry such that:

`1 - sin^2 t = cos^2 t`

`100 int sqrt(1 - sin^2 t)*cos t dt = 100 int sqrt(cos^2 t)*cos t dt`

`100 int sqrt(1 - sin^2 t)*cos t dt = 100 int cos^2 t dt`

You need to use the half angle trigonmetric identity such that:

`cos^2 t = (1 + cos 2t)/2`

`100 int cos^2 t dt = 100 int(1 + cos 2t)/2 dt`

Using the linearity of integral yields:

`100 int (1 + cos 2t)/2 dt = 100 int (1/2) dt+ 100 int (cos 2t)/2 dt`

`100 int (1 + cos 2t)/2 dt = 50 t + 25 sin 2t + c`

Substituting back `sin^(-1)((x-10)/10)` for `t` yields:

`int sqrt(20x-x^2)dx = 50sin^(-1)((x-10)/10) + 25 sin 2(sin^(-1)((x-10)/10)) + c`

`int sqrt(20x-x^2)dx = 50sin^(-1)((x-10)/10) + 50 sin(sin^(-1)((x-10)/10))cos(sin^(-1)((x-10)/10)) + c`

`int sqrt(20x-x^2)dx = 50sin^(-1)((x-10)/10) + 5(x-10)*sqrt(1 - (x-10)^2/100) + c`

**Hence, evaluating the given integral, using trigonometric substitution yields** `int sqrt(20x-x^2)dx = 50sin^(-1)((x-10)/10) + 5(x-10)*sqrt(1 - (x-10)^2/100) + c.`

Integrate the original integrand by trigonometric substitution:

∫ √(20x - x²) dx = ∫ √[-(x² - 20x)] dx

∫ √(20x - x²) dx = ∫ √[-{(x - 10)² - 100}] dx

∫ √(20x - x²) dx = ∫ √[100 - (x - 10)²] dx

Let x - 10 = 10sinθ,

x = 10 + 10sinθ

dx / dθ = 10cosθ

dx = 10cosθ dθ

∫ √(20x - x²) dx = ∫ 10cosθ√[100 - (10sinθ)²] dθ

∫ √(20x - x²) dx = ∫ 10cosθ√(100 - 100sin²θ) dθ

∫ √(20x - x²) dx = ∫ 10cosθ√[100(1 - sin²θ)] dθ

∫ √(20x - x²) dx = ∫ 10cosθ√(100cos²θ) dθ

∫ √(20x - x²) dx = ∫ 10cosθ × 10cosθ dθ

∫ √(20x - x²) dx = 100 ∫ cos²θ dθ

∫ √(20x - x²) dx = 50 ∫ [1 + cos(2θ)] dθ

∫ √(20x - x²) dx = 50θ + 25sin(2θ)

Since θ = sinˉ¹[(x - 10) / 10],

∫ √(20x - x²) dx = 50sinˉ¹[(x - 10) / 10] + 25sin[2sinˉ¹{(x - 10) / 10}]

∫ √(20x - x²) dx = 50sinˉ¹[(x - 10) / 10] + 50sin[sinˉ¹{(x - 10) / 10}]cos[sinˉ¹{(x - 10) / 10}]

∫ √(20x - x²) dx = 50sinˉ¹[(x - 10) / 10] + (x - 10)√(20x - x²) / 2

∫ √(20x - x²) dx = (x - 10)√(20x - x²) / 2 + 50sinˉ¹[(x - 10) / 10] + C