# Evaluate the indefinite integral integrate of sqrt(20x-x^2)dx All previous answers to this question have been incorrect.

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You should complete the square using the formula `(a-b)^2 = a^2 - 2ab + b^2` , such that:

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

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

You need to factor out 100 such that:

`sqrt(100(1 - ((x - 10)/10)^2))`

You need to use the following trigonometric substitution such that:

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

You need to change the variable such that:

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

Using the fundamental formula of trigonometry yields:

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

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

`int 100(sqrt(cos^2 t))(cos t dt) = int 100(cos t)(cos t dt)`

`int 100(cos t)(cos t dt) = 100 int cos^2 t dt`

You should use the following trigonometric 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 property of linearity yields:

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

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

Substituting back `arcsin((x-10)/10)` for t yields:

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

**Hence, evaluating the given integral, using trigonometric substitution yields `int sqrt(20x-x^2) dx = 50arcsin((x-10)/10) + 25(sin 2arcsin((x-10)/10)) + c.` **

You might want to be careful. The answers to complicated integrals can often appear in many different guises. If the solution presented does not match a textbook solution or solution by a computer algebra system, it may be because a different substitution was made. Both answers may be correct and yet look very different. The way to check is to (correctly) take the derivative.

Evaluate `int sqrt(20x-x^2)dx` Complete the square in the radicand:

`=int sqrt(100-(x-10)^2)dx` Let `u=x-10,du=dx`

`=int sqrt(100-u^2)du`

Let `u=10sin(s),du=10cos(s)ds` . Now `sqrt(100-u^2)=sqrt(100-100sin^2(s))=sqrt(100(1-sin^2(s)))=10cos(s)` so`sqrt(100-u^2)du=100cos^2(s)ds`

`=100 int cos^2(s)ds` With `cos^2(s)=1/2cos(2s)+1/2` :

`=50intcos(2s)ds+50 int ds` Let `p=2s,dp=2ds`

`=25int cos(p) dp+50int ds`

`=25sin(p)+50 int ds+C_1`

`=25sin(2s)+50s+C_2` But `u=10sin(s)` so `s=sin^(-1)(u/10)`

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```25sin(2s)=50sin(s)cos(s)=5sin(s)10sqrt(1-sin^2(s))`

`=5sin(s)sqrt(100(1-sin^2(s)))=5sin(s)sqrt(100-100sin^2(s))`

But `u^2=100sin^2(s)` so `25sin(2s)=1/2usqrt(100-u^2)`

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`=1/2usqrt(100-u^2)+50sin^(-1)(u/10)+C_2`

Substituting for `u` we get

`=1/2(x-10)sqrt(100-(x-10)^2)+50sin^(-1)((x-10)/10)+C` or

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`intsqrt(20x-x^2)dx = 1/2(x-10)sqrt(20x-x^2)+50sin^(-1)(x/10-1)+C`

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There are a number of ways to rewrite this answer. For example you might write it as `1/2[(x-10)sqrt(20x-x^2)-100sin^(-1)(1-x/10)]+C` or any of a number of equivalent expressions.