Question

solution: (d); \\(\int \frac{x}{\sqrt{1 - x^2}} \\, dx = -\frac{1}{2} \int \frac{-2x}{\sqrt{1 - x^2}} \\, dx\\)\\(= -\frac{1}{2} \times 2\sqrt{1 - x^2} + c = -\sqrt{1 - x^2} + c\\)

Explanation:

Step1: Recognize substitution form

The integral \(\int \frac{x}{\sqrt{1 - x^2}} dx\) can be handled by noticing that the derivative of \(1 - x^2\) is \(-2x\), so we manipulate the integrand to match the form of the derivative of the inner function. We rewrite the integrand as \(-\frac{1}{2}\int \frac{-2x}{\sqrt{1 - x^2}} dx\).

Step2: Apply substitution

Let \(u = 1 - x^2\), then \(du = -2x dx\). So the integral becomes \(-\frac{1}{2}\int \frac{du}{\sqrt{u}}\). The integral of \(\frac{1}{\sqrt{u}}\) is \(2\sqrt{u}+C\), so multiplying by \(-\frac{1}{2}\) gives \(-\frac{1}{2}\times2\sqrt{1 - x^2}+c\).

Step3: Simplify the expression

Simplifying \(-\frac{1}{2}\times2\sqrt{1 - x^2}+c\) gives \(-\sqrt{1 - x^2}+c\).

Answer:

The integral \(\int \frac{x}{\sqrt{1 - x^2}} dx\) evaluates to \(-\sqrt{1 - x^2}+c\) (where \(c\) is the constant of integration).