5.5 The Substitution Rule/41: Difference between revisions
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<math> | <math> | ||
\int \frac{1}{\sqrt{1-x^{2}} \arcsin {x}} | \int \frac{1}{\sqrt{1-x^{2}} \arcsin {x}} | ||
</math> | </math> | ||
| Line 10: | Line 10: | ||
u &= \arcsin {x} | u &= \arcsin {x} | ||
\\[2ex] | \\[2ex] | ||
du &= \frac{1}{\sqrt{1-x^2}} dx \\[2ex] | du &= \frac{1}{\sqrt{1-x^2}}\;dx \\[2ex] | ||
\end{align} | \end{align} | ||
| Line 18: | Line 18: | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
\int \frac{1}{u} du | \int (\frac{1}{\sqrt{1-x^{2}}}\frac{1}{\arcsin {x}})\;dx | ||
&= \ln |u| +c [2ex] | &= \int \frac{1}{u}\;du \\[2ex] | ||
&= \ln |\arcsin {x}| + c [2ex] | &= \ln |u| +c \\[2ex] | ||
&= \ln |\arcsin {x}| + c \\[2ex] | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Latest revision as of 23:19, 13 September 2022
![{\displaystyle {\begin{aligned}u&=\arcsin {x}\\[2ex]du&={\frac {1}{\sqrt {1-x^{2}}}}\;dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/b4508e836235d223e2f8c84c555442628b62e838)
![{\displaystyle {\begin{aligned}\int ({\frac {1}{\sqrt {1-x^{2}}}}{\frac {1}{\arcsin {x}}})\;dx&=\int {\frac {1}{u}}\;du\\[2ex]&=\ln |u|+c\\[2ex]&=\ln |\arcsin {x}|+c\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/dc486b5a2bab32ba3d08998cd31ea83afe03b4c6)