Student Question

`int arcsin(4x) dx` Use integration tables to find the indefinite integral.

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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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