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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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.`

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akshaygoel96 | Student, Grade 11 | (Level 1) Honors

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There is no need to add a constant before completing the square as some answerers have done, you can complete the square straight away.

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
 
this will help you.....(:

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