5.5 The Substitution Rule/41: Difference between revisions
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\int \frac{1}{\sqrt{1-x^{2}}} = \int \frac{1}{u} du = \ln |u| +c = \ln |\arcsin {x}| + c | \int \frac{1}{\sqrt{1-x^{2}}} = \int \frac{1}{u} du = \ln |u| +c = \ln |\arcsin {x}| + c | ||
</math> | |||
<math> | |||
\begin{align} | |||
u &= \sqrt{u} | |||
\\[2ex] | |||
du &= \frac{1}{2}\ \frac{1}{\sqrt{t}} dx \\[2ex] | |||
2du &= \frac{1}{\sqrt{t}} dx | |||
\end{align} | |||
</math> | </math> | ||
Revision as of 09:06, 7 September 2022
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int \frac{1}{\sqrt{1-x^{2}}} = \int \frac{1}{u} du = \ln |u| +c = \ln |\arcsin {x}| + c }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} u &= \sqrt{u} \\[2ex] du &= \frac{1}{2}\ \frac{1}{\sqrt{t}} dx \\[2ex] 2du &= \frac{1}{\sqrt{t}} dx \end{align} }