lunnatican4

2022-01-05

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

Papilys3q

Step 1: Given that
Evaluate the following integral:
$\int \frac{1-x}{\sqrt{1-{x}^{2}}}dx$
Step 2: Solve
We have,
$\int \frac{1-x}{\sqrt{1-{x}^{2}}}dx=\int \frac{1}{\sqrt{1-{x}^{2}}}dx-\int \frac{x}{\sqrt{1-{x}^{2}}}dx$
$={\mathrm{sin}}^{-1}\left(x\right)-\int \frac{x}{\sqrt{t}}×\frac{dt}{-2x}$
$={\mathrm{sin}}^{-1}x+\frac{1}{2}\left[\frac{{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]+C$
$={\mathrm{sin}}^{-1}x+\sqrt{t}+C$
$={\mathrm{sin}}^{-1}x+\sqrt{1-{x}^{2}}+C$

Robert Pina

Given:
$\int \frac{1-x}{\sqrt{1-{x}^{2}}}dx$
$\int \frac{1}{\sqrt{1-{x}^{2}}}-\frac{x}{\sqrt{1-{x}^{2}}}dx$
Use properties
$\int \frac{1}{\sqrt{1-{x}^{2}}}dx-\int \frac{x}{\sqrt{1-{x}^{2}}}dx$
$\mathrm{arcsin}\left(x\right)+\sqrt{1-{x}^{2}}$
Solution:
$\mathrm{arcsin}\left(x\right)+\sqrt{1-{x}^{2}}+C$

karton

$\int \frac{1-x}{\sqrt{1-{x}^{2}}}dx\phantom{\rule{0ex}{0ex}}=\int \left(\frac{1}{\sqrt{1-{x}^{2}}}-\frac{x}{\sqrt{1-{x}^{2}}}\right)dx\phantom{\rule{0ex}{0ex}}=\int \frac{1}{\sqrt{1-{x}^{2}}}dx-\int \frac{x}{\sqrt{1-{x}^{2}}}dx\phantom{\rule{0ex}{0ex}}\text{Now calculate}\phantom{\rule{0ex}{0ex}}\int \frac{1}{\sqrt{1-{x}^{2}}}dx\phantom{\rule{0ex}{0ex}}=\mathrm{arcsin}\left(x\right)\phantom{\rule{0ex}{0ex}}\text{Now calculate}\phantom{\rule{0ex}{0ex}}\int \frac{x}{\sqrt{1-{x}^{2}}}dx\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}\int \frac{1}{\sqrt{u}}du\phantom{\rule{0ex}{0ex}}\int \frac{1}{\sqrt{u}}du\phantom{\rule{0ex}{0ex}}=2\sqrt{u}\phantom{\rule{0ex}{0ex}}-\frac{1}{2}\int \frac{1}{\sqrt{u}}du\phantom{\rule{0ex}{0ex}}=-\sqrt{u}\phantom{\rule{0ex}{0ex}}=-\sqrt{1-{x}^{2}}\phantom{\rule{0ex}{0ex}}\int \frac{1}{\sqrt{1-{x}^{2}}}dx-\int \frac{x}{\sqrt{1-{x}^{2}}}dx\phantom{\rule{0ex}{0ex}}=\mathrm{arcsin}\left(x\right)+\sqrt{1-{x}^{2}}\phantom{\rule{0ex}{0ex}}=\mathrm{arcsin}\left(x\right)+\sqrt{1-{x}^{2}}+C$