Evaluate the integral? : # int_0^2 xsqrt(2x-x^2) dx #

Consider the indefinite integral:

# I = int \ xsqrt(2x-x^2) \ dx #

Which we can write as:

# I = int \ (x-1)sqrt(2x-x^2) + sqrt(2x-x^2) \ dx #
# \ \ = int \ (x-1)sqrt(2x-x^2) \ dx + int \ sqrt(2x-x^2) \ dx #

For the first integral, we have:

# I_1 = int \ (x-1)sqrt(2x-x^2) \ dx #

we can use a substitution, Let:

# u = 2x-x^2 => (du)/dx = 2-2x = -2(x-1) #

And if we perform this substitution then we get:

# I_1 = int \ (-1/2) sqrt(u) \ du #
# \ \ \ = -1/2 u^(3/2)/(3/2) #
# \ \ \ = -1/3 u^(3/2) #

And restoring the substitution we get:

# I_1 = -1/3 (2x-x^2)^(3/2) #

Next we consider the second integral,

# I_2 = int \ sqrt(2x-x^2) \ dx #
# \ \ \ = int \ sqrt(1-(x-1)^2) \ dx #

And here we can perform a substitution; Let

# sin theta=x-1 => (d theta)/(dx)cos theta=1 #

And if we perform this substitution then we get:

# I_2 = int \ sqrt(1-sin^2 theta) \ (cos theta) \ d theta #
# \ \ \ = int \ cos^2 theta \ d theta #
# \ \ \ = int \ (cos(2theta)+1)/2 \ d theta #
# \ \ \ = (sin2theta + theta)/2 #

And restoring the substitution we find that # theta = arcsin(x-1)# and #sin(2theta)=2(x-1)sqrt(1-(x-1)^2) # and so:

# I_2 = (arcsin(x-1) + (x-1)sqrt((x-1)-(x-1)^2))/2 #
# \ \ \ = (arcsin(x-1) + (x-1)sqrt(2x-x^2))/2 #

Combining our two results we then gave:

# I = (arcsin(x-1) + (x-1)sqrt(2x-x^2))/2 -1/3 (2x-x^2)^(3/2)#

Given this result we then have:

# int_0^2 \ xsqrt(2x-x^2) \ dx = [(arcsin(x-1) + (x-1)sqrt(2x-x^2))/2 -1/3 (2x-x^2)^(3/2)]_0^2 #

# " " = 1/2((arcsin1 - arcsin(-1)) #

# " " = 1/2(pi/2-(-pi/2)) #

# " " = pi/2 #

#int_0^2 xsqrt(2x-x^2)*dx#

=#int_0^2 xsqrt(1^2-(x-1)^2)*dx#

After using #x-1=sinu#, #x=1+sinu# and #dx=cosu*du# transforms, this integral became

#int_(-pi/2)^(pi/2) (1+sinu)*cosu*cosu*du#

=#int_(-pi/2)^(pi/2) (1+sinu)*(cosu)^2*du#

=#1/2int_(-pi/2)^(pi/2) (1+sinu)*(1+cos2u)*du#

=#1/2int_(-pi/2)^(pi/2) (1+cos2u+sinu+cos2u*sinu)*du#

=#1/2int_(-pi/2)^(pi/2) [1+cos2u+sinu+1/2*(sin3u-sinu)]*du#

=#1/4int_(-pi/2)^(pi/2) (2+2cos2u+sin3u+sinu)*du#

=#1/4*[2u+sin2u-1/3*cos3u-cosu]_(-pi/2)^(pi/2)#

=#1/4*2pi#

=#pi/2#