Rewrite the integrand as follows.
int xsqrt(4x-x^2) "d"x = int xsqrt(4-(x-2)^2)"d"x∫x√4x−x2dx=∫x√4−(x−2)2dx.
Denote the required integral int xsqrt(4x-x^2) "d"x∫x√4x−x2dx by II. Make the substitution u=x-2u=x−2. Then II is transformed to,
I = int (u+2) sqrt(4-u^2) "d"uI=∫(u+2)√4−u2du.
I = int usqrt(4-u^2) "d"u + 2int sqrt(4-u^2) "d"uI=∫u√4−u2du+2∫√4−u2du.
A good general integration result to be aware of is, for n != -1n≠−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.
This can help us solve the first integral int usqrt(4-u^2) "d"u as picking 4-u^2=p(x) gives p'(x)=-2u. We conclude, int u(4-u^2)^(1/2) "d"u = -1/3 (4-u^2)^(3/2) + C. Then,
I = -1/3 (4-u^2)^(3/2) + 2intsqrt(4-u^2) "d"u.
Denote J=int sqrt(4-u^2) "d"u. Make the substitution u=2sin(theta). Then "d"u = 2cos(theta) "d"theta. J is transformed to,
J = 2int sqrt(4(1-sin^2(theta)))cos(theta) "d"theta.
The trigonometric identity cos^2(theta)+sin^2(theta)=1 gives sqrt(1-sin^2(theta))=cos^2(theta).
J = 4 int cos^2(theta) "d"theta.
The double angle identity cos(2theta) = 2cos^2(theta)-1 gives cos^2(theta)=1/2(1+cos(2theta)). Then,
J = 2 int 1+cos(2theta) "d"theta
J = 2theta + sin(2theta) + C
J = 2theta + 2sin(theta)cos(theta) + C
J = 2theta + 2sin(theta)sqrt(1-sin^2(theta))+C.
Then u=2sin(theta) gives theta = "arcsin"(u/2).
Then,
J = 2"arcsin"(u/2) + usqrt(1-u^2/4)+C
J = 2"arcsin"(u/2) + u/2sqrt(4-u^2)+C
We conclude,
I = 4arcsin(u/2) + usqrt(4-u^2) - 1/3(4-u^2)^(3/2) + C.
I = 4arcsin(u/2) + 1/3(u^2+3u-4)sqrt(4-u^2) + C.
Then u=x-2. u^2=x^2-4x+4. Gives u^2+3u-4=x^2-x-6.
Then,
I = 2arcsin((x-2)/2) + 1/3 (x^2-x-6) sqrt(4x-x^2) + C.
