Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 87

For the given differential equation: (dy)/(dx) = 1/sqrt(80+8x-16x^2), we may write it in a form of N(y) dy = M(x) dx .
Cross-multiply the (dx) to the other side:
(dy) = 1/sqrt(80+8x-16x^2) dx
To solve for the general solution of the differential equation, we may apply direct integration on both sides.
int (dy) = int 1/sqrt(80+8x-16x^2) dx
For the left side, it follow basic integral formula:
int dy = y
To evaluate the right side, we may apply completing the square on the trinomial: 80+8x-16x^2 = -(4x-1)^2+81 or 81-(4x-1)^2
Then, the integral on the right side becomes:
int 1/sqrt(80+8x-16x^2) dx=int 1/sqrt(81-(4x-1)^2) dx
The integral resembles the basic integration formula for inverse sine function:
int 1/sqrt(a^2-u^2)du=arcsin(u/a)+C
We let u = 4x-1 then du = 4 dx or (du)/4= dx .
Note that 81 = 9^2
Then,
int 1/sqrt(81-(4x-1)^2) dx =int 1/sqrt(9^2-u^2) *(du)/4
=(1/4)int1/sqrt(9^2-u^2)du
=(1/4)arcsin(u/9)+C
Plug-in u=4x-1 in (1/4)arcsin(u/9 ), we get:
int 1/sqrt(81-(4x-1)^2) dx= (1/4)arcsin((4x-1)/9) +C
Combining the results from both sides, we get the general solution of differential equation:
y =1/4arcsin((4x-1)/9) +C