Question

evaluate the integral.
\\( \int \frac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}} dx \\)
\\( \int \frac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}} dx = \square \\)

Explanation:

Step1: Identify substitution

Let \( u = \sin^{-1} x \). Then, find \( du \). The derivative of \( \sin^{-1} x \) with respect to \( x \) is \( \frac{1}{\sqrt{1 - x^2}} \), so \( du = \frac{1}{\sqrt{1 - x^2}} dx \).

Step2: Substitute into integral

The integral \( \int \frac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}} dx \) becomes \( \int e^u du \) after substituting \( u \) and \( du \).

Step3: Integrate \( e^u \)

The integral of \( e^u \) with respect to \( u \) is \( e^u + C \), where \( C \) is the constant of integration.

Step4: Substitute back \( u \)

Substitute \( u = \sin^{-1} x \) back into the result. So we get \( e^{\sin^{-1} x} + C \).

Answer:

\( e^{\sin^{-1} x} + C \) (where \( C \) is the constant of integration)