5.5 The Substitution Rule/13: Difference between revisions

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(Created page with "<math> \int \frac{\sin{(\ln{(x))}}}{x}dx </math> <math> \begin{align} u &=\ln(x) \\[2ex] du &= \frac{1}{x}dx \\[2ex] \end{align} </math> <math> \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} </math>")
 
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<math>
\int \frac{\sin{(\ln{(x))}}}{x}dx
</math>


<math>
\begin{align}
u &=\ln(x) \\[2ex]
du &= \frac{1}{x}dx \\[2ex]
\end{align}
</math>
<math>
\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}
</math>

Revision as of 22:45, 13 September 2022

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