# What is the integration of `intd/dx e^arcsin(x^2) dx `

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### 1 Answer

You need to differentiate the given function `y = e^(arcsin(x^2))` with respect to x, such that:

`(d(e^(arcsin(x^2))))/dx = e^(arcsin(x^2))*(2x)/(sqrt(1 - x^4))`

You need to come up with the substitution,such that:

`x^2 = t => 2xdx = dt`

Changing the variable yields:

`int (d(e^(arcsin(x^2))))/dx dx = int e^(arcsin(x^2))*(2x)/(sqrt(1 - x^4)) dx`

`int e^(arcsin(x^2))*(2x)/(sqrt(1 - x^4)) dx = int e^(arcsin t)*(dt)/(sqrt(1 - t^2))`

You should come up with the next substitution, such that:

`arcsin t = u => (dt)/(sqrt(1 - t^2)) = du`

Changing the variable again yields:

`int e^(arcsin t)*(dt)/(sqrt(1 - t^2)) = int e^u du = e^u + c`

Substituting back `arcsin t` for u, yields:

`int e^(arcsin t)*(dt)/(sqrt(1 - t^2)) = e^(arcsin t) + c`

Substituting back `x^2` for t yields:

`int e^(arcsin(x^2))*(2x)/(sqrt(1 - x^4)) dx = e^(arcsin x^2) + c `

**Hence, evaluating the given indefinite integral, yields **`int (d(e^(arcsin(x^2))))/dx dx = = e^(arcsin x^2) + c.`