Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 18

From the table of integrals, we have a integration formula for inverse sine function as:
int arcsin(u/a)du = u*arcsin(u/a) +sqrt(a^2-u^2) +C
It resembles the given integral problem: int arcsin(4x)dx or int arcsin((4x)/1)dx where u =4x and a=1 ,
When we let u = 4x , we solve for the derivative of "u" as: du = 4 dx or (du)/4= dx .
Plug-in u = 4x and (du)/4=dx on the integral problem, we get:
int arcsin(4x)dx =int arcsin(u) * (du)/4
Apply the basic properties of integration: int c*f(x) dx= c int f(x) dx .
int arcsin(u) * (du)/4 = 1/4int arcsin(u) du or 1/4int arcsin(u/1) du
Applying the integral formula for inverse sine function, we get:
1/4 int arcsin(u/1)du = (1/4) *[u*arcsin(u/1) +sqrt(1^2-u^2)] +C
= (1/4) *[u*arcsin(u) +sqrt(1-u^2)] +C
= (u*arcsin(u))/4 +sqrt(1-u^2)/4 +C
Plug-in u =4x on (u*arcsin(u))/4 +sqrt(1-u^2)/4 +C , we get indefinite integral as:
int arcsin(4x)dx =(4x*arcsin(4x))/4 +sqrt(1-(4x)^2)/4 +C
=(4x*arcsin(4x))/4 +sqrt(1-16x^2)/4 +C
= x*arcsin(4x) +sqrt(1-16x^2)/4 +C