Evaluate the integral #int \ sqrt(x-x^2)/x \ dx #?

Question

Aurora Alvarado · Accepted Answer

#int sqrt(x-x^2)/x "d"x =xsqrt(1/x-1)-arctan(sqrt(1/x-1))+C#.

Note that the integrand requires #x>0#. For #x>0#, #x=sqrt(x^2)#. Then rewrite the integrand as, #sqrt(x-x^2)/x=sqrt((x-x^2)/x^2)#. #sqrt(x-x^2)/x=sqrt(1/x-1)#. Then denote the required integral by #I#. #I=int sqrt(1/x-1) "d"x#. Make the substitution #u=sqrt(1/x-1)#. Then #"d"x = -2x^2sqrt(1/x-1)"d"u#. Then #I# is transformed to, #I = -2 int x^2 (1/x-1) "d"u#. From the definition of #u#, #1/x-1=u^2#. Then #x=1/(1+u^2)#, #x^2=1/(1+u^2)^2#. Then, #I = -2 int u^2/(u^2+1)^2 "d"u#, #I = - int u * (2u)/(u^2+1)^2 "d"u#. Integration by parts states #int f'(u)g(u) "d"u = f(u)g(u) - int f(u)g'(u)"d"u#. Let #f'(u)= 2u/(u^2+1)^2#, #g(u)=u#. A good general integration result to be aware of is, for #n != -1#, #int p'(x) [p(x)]^(n) "d"x = ([p(x)]^(n+1))/(n+1)+C#. This can be easily verified by differentiating the LHS. Then as #f'(u)# is in this form we conclude #f(u) = -1/(u^2+1)#. Then, by parts, #I = -(-1/(u^2+1)*u - int -1/(u^2+1)"d"u)#, #I = u/(u^2+1) - int 1/(u^2+1) "d"u#, Another general integration result to be aware of is #int 1/(u^2+1) "d"u = arctan(u) + C#. Then, #I = u/(u^2+1) - arctan(u)+C#. We know #u=sqrt(1/x-1)#. Then #u^2+1=1/x#. We conclude, #I=xsqrt(1/x-1)-arctan(sqrt(1/x-1))+C#.

Aurora Alvarado · Answer

# int \ sqrt(x-x^2)/x \ dx = 1/2 arcsin(2x-1)+sqrt(x - x^2) + C#

We seek: # I = int \ sqrt(x-x^2)/x \ dx # We can start by completing the square of the numerator to get: # x-x^2 = -(x^2-x) # # " " = -{(x-1/2)^2-(1/2)^2} # # " " = 1/4-(x-1/2)^2 # # " " = 1/4{1-(2x-1)^2} # So we can write: # I = int \ sqrt(1/4{1-(2x-1)^2})/x \ dx # # \ \ = 1/2 \ int \ sqrt(1-(2x-1)^2)/x \ dx # Then comparing the numerator with #1-sin^2A# we could try a trig substitution of the form: # sintheta = 2x-1 => cos theta (d theta)/dx = 2#, and #x=1/2(sin theta+1) # Substituting this into the integral and we get: # I = 1/2 \ int \ sqrt(1-sin^2 theta)/(1/2(sin theta+1)) \ 1/2cos theta \ d theta # # \ \ = 1/2 \ int \ cos^2 theta/(sin theta+1) \ d theta # # \ \ = 1/2 \ int \ (1-sin^2 theta)/(1+sin theta) \ d theta # # \ \ = 1/2 \ int \ ((1+sin theta)(1-sin theta))/(1+sin theta) \ d theta # # \ \ = 1/2 \ int \ 1-sin theta \ d theta # # \ \ = 1/2 \ {theta+cos theta} + C# And from the substitution we have: # sin^2theta = (2x-1)^2 => cos^2 theta = 1 - (2x-1)^2 # # :. cos^2 theta = 1 - 4x^2+4x-1 # # " " = 4x - 4x^2 # And so we can reverse the substitution to get: # I = 1/2 \ {arcsin(2x-1)+sqrt(4(x - x^2))} + C# # \ \ = 1/2 arcsin(2x-1)+sqrt(x - x^2) + C#

Adam Adkins · Answer

undefined The integral of $\int \frac{\sqrt{x-x^2}}{x} \, dx$ can be evaluated as follows:

$$ \int \frac{\sqrt{x-x^2}}{x} \, dx = \frac{\pi}{2} $$