# Evaluate the following integral: \int \frac{1-x}{\sqrt{1-x^{2}}}dx

Evaluate the following integral:
$$\displaystyle\int{\frac{{{1}-{x}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Papilys3q
Step 1: Given that
Evaluate the following integral:
$$\displaystyle\int{\frac{{{1}-{x}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}$$
Step 2: Solve
We have,
$$\displaystyle\int{\frac{{{1}-{x}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}=\int{\frac{{{1}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}-\int{\frac{{{x}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={{\sin}^{{-{1}}}{\left({x}\right)}}-\int{\frac{{{x}}}{{\sqrt{{{t}}}}}}\times{\frac{{{\left.{d}{t}\right.}}}{{-{2}{x}}}}$$
$$\displaystyle={{\sin}^{{-{1}}}{x}}+{\frac{{{1}}}{{{2}}}}{\left[{\frac{{{t}^{{-{\frac{{{1}}}{{{2}}}}+{1}}}}}{{-{\frac{{{1}}}{{{2}}}}+{1}}}}\right]}+{C}$$
$$\displaystyle={{\sin}^{{-{1}}}{x}}+\sqrt{{{t}}}+{C}$$
$$\displaystyle={{\sin}^{{-{1}}}{x}}+\sqrt{{{1}-{x}^{{{2}}}}}+{C}$$
###### Not exactly what you’re looking for?
Robert Pina
Given:
$$\displaystyle\int{\frac{{{1}-{x}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle\int{\frac{{{1}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}-{\frac{{{x}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}$$
Use properties
$$\displaystyle\int{\frac{{{1}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}-\int{\frac{{{x}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle{\arcsin{{\left({x}\right)}}}+\sqrt{{{1}-{x}^{{{2}}}}}$$
Solution:
$$\displaystyle{\arcsin{{\left({x}\right)}}}+\sqrt{{{1}-{x}^{{{2}}}}}+{C}$$
karton

$$\int \frac{1-x}{\sqrt{1-x^{2}}}dx \\=\int (\frac{1}{\sqrt{1-x^{2}}}-\frac{x}{\sqrt{1-x^{2}}})dx \\=\int \frac{1}{\sqrt{1-x^{2}}}dx-\int \frac{x}{\sqrt{1-x^{2}}}dx \\\text{Now calculate} \\\int \frac{1}{\sqrt{1-x^{2}}}dx \\=\arcsin(x) \\\text{Now calculate} \\\int \frac{x}{\sqrt{1-x^{2}}}dx \\=-\frac{1}{2}\int \frac{1}{\sqrt{u}}du \\\int \frac{1}{\sqrt{u}}du \\=2\sqrt{u} \\-\frac{1}{2}\int \frac{1}{\sqrt{u}}du \\=-\sqrt{u} \\=-\sqrt{1-x^{2}} \\\int \frac{1}{\sqrt{1-x^{2}}}dx-\int \frac{x}{\sqrt{1-x^{2}}}dx \\=\arcsin(x)+\sqrt{1-x^{2}} \\=\arcsin(x)+\sqrt{1-x^{2}}+C$$