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

lunnatican4 2022-01-05 Answered
Evaluate the following integral:
\(\displaystyle\int{\frac{{{1}-{x}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}\)

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Expert Answer

Papilys3q
Answered 2022-01-06 Author has 2940 answers
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,
\(\displaystyle\int{\frac{{{1}-{x}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}=\int{\frac{{{1}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}-\int{\frac{{{x}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={{\sin}^{{-{1}}}{\left({x}\right)}}-\int{\frac{{{x}}}{{\sqrt{{{t}}}}}}\times{\frac{{{\left.{d}{t}\right.}}}{{-{2}{x}}}}\)
\(\displaystyle={{\sin}^{{-{1}}}{x}}+{\frac{{{1}}}{{{2}}}}{\left[{\frac{{{t}^{{-{\frac{{{1}}}{{{2}}}}+{1}}}}}{{-{\frac{{{1}}}{{{2}}}}+{1}}}}\right]}+{C}\)
\(\displaystyle={{\sin}^{{-{1}}}{x}}+\sqrt{{{t}}}+{C}\)
\(\displaystyle={{\sin}^{{-{1}}}{x}}+\sqrt{{{1}-{x}^{{{2}}}}}+{C}\)
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Robert Pina
Answered 2022-01-07 Author has 1798 answers
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:
\(\displaystyle{\arcsin{{\left({x}\right)}}}+\sqrt{{{1}-{x}^{{{2}}}}}+{C}\)
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karton
Answered 2022-01-11 Author has 8659 answers

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

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