5.5 The Substitution Rule/9: Difference between revisions

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\begin{align}
\begin{align}


u &=\ln(x) \\[2ex]
u &=3x-2 \\[2ex]
du &= \frac{1}{x}dx \\[2ex]
du &= 3dx \\[2ex]
\frac{1}{3}du &= dx \\[2ex]
 
\end{align}
</math>
 
<math>
\begin{align}
 
\int(u)^{20}/frac{1}{3}du \\[2ex]
 
&= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex]
&= -\cos{(u)} + C \\[2ex]
&= -\cos{(\ln{(x)})} + C


\end{align}
\end{align}
</math>
</math>

Revision as of 03:53, 4 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 (3x-2)^{20} dx }


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 &=3x-2 \\[2ex] du &= 3dx \\[2ex] \frac{1}{3}du &= dx \\[2ex] \end{align} }

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} \int(u)^{20}/frac{1}{3}du \\[2ex] &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] &= -\cos{(u)} + C \\[2ex] &= -\cos{(\ln{(x)})} + C \end{align} }