5.5 The Substitution Rule/30: Difference between revisions
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&= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] | &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] | ||
&= -\cos{(u)} + C \\[2ex] | &= -\cos{(u)} + C \\[2ex] | ||
&= -\cos{(\ln{(x)}} + C | &= -\cos{(\ln{(x)})} + C | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 19:08, 26 August 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{\sin{(\ln{(x))}}}{x}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 &=\ln(x) \\[2ex] du &= \frac{1}{x}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 \frac{\sin{(\ln{(x))}}}{x}dx &= \int\frac{1}{x}\sin(\ln{(x)})dx = \int\left(\frac{1}{x}dx\right)\sin{(\ln{(x)})} \\[2ex] &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] &= -\cos{(u)} + C \\[2ex] &= -\cos{(\ln{(x)})} + C \end{align} }