5.5 The Substitution Rule/54: Difference between revisions

From Mr. V Wiki Math
Jump to navigation Jump to search
No edit summary
No edit summary
Line 19: Line 19:
\begin{align}
\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_{0}^{7} \sqrt{4+3x}\,dx \\[2ex]


&= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex]
&= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex]

Revision as of 19:16, 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_{0}^{7} \sqrt{4+3x}\,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 &=4+3x \\[2ex] du &= 3\,dx \\[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_{0}^{7} \sqrt{4+3x}\,dx \\[2ex] &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] &= -\cos{(u)} + C \\[2ex] &= -\cos{(\ln{(x)})} + C \end{align} }