Question

Integration problem​

Answer 1

`e^{\sin^(-1)(x)}+C`

Explanation:

To solve this integral, we will consider the properties of the inverse sine function and its relation to the expression √(1 - x²) in the denominator, which is reminiscent of the derivative of the inverse sine function. By recognizing this relationship, we can find the integral of the given function.

Given integral:

`\displaystyle \int \Big(\frac{e^{\sin^(-1)(x)}}{√(1-x^2)}\Big)dx`

`\hrulefill`

Using u-substitution, let u = sin⁻¹(x) then du = 1/√(1-x²) · dx. Make these substitutions:

`\text{We have, } \\ \\ \displaystyle \int \Big(\frac{e^{\sin^(-1)(x)}}{√(1-x^2)}\Big)dx \\ \\ \\ \text{Letting, } \\\\ u=\sin^(-1)(x) \rightarrow du=(1)/(√(1-x^2))dx \\ \\ \\ \\ \Longrightarrow \int \left((e^u)/(√(1-x^2))\right)√(1-x^2)du\\\\\\\\`

With these substitutions, our integral becomes:

`\displaystyle \Longrightarrow \int e^udu`

Which is straightforward to integrate. Let's perform the computation. The integral of e^u with respect to u is simply e^u:

`\Longrightarrow e^u + C`

Now, we need to substitute back using u = sin⁻¹(x) to express our result in terms of 'x'.

`\therefore \displaystyle \int \Big(\frac{e^{\sin^(-1)(x)}}{√(1-x^2)}\Big)dx = \boxed{e^{\sin^(-1)(x)}+C}`

Thus, the integral is solved where 'C' is the constant of integration.