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

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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.`

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

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)`

------------------------------------------------------

```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)`

---------------------------------------------------------

`=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

***************************************************

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

***************************************************

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.