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`
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