5.5 The Substitution Rule/54: Difference between revisions
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\begin{align} | \begin{align} | ||
\int_{0}^{ | \int_{0}^{\sqrt{\pi}} x\cos{(x^2)}\,dx = \int_{0}^{\sqrt{\pi}} (xdx)\cos{(x^2)} = | ||
&= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] | &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] | ||
Revision as of 19:23, 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}^{\sqrt{\pi}} x\cos{(x^2)}\,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 &=x^2 \\[2ex] du &= 2xdx \\[2ex] \frac{1}{2}du &= xdx \\[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}^{\sqrt{\pi}} x\cos{(x^2)}\,dx = \int_{0}^{\sqrt{\pi}} (xdx)\cos{(x^2)} = &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] &= -\cos{(u)} + C \\[2ex] &= -\cos{(\ln{(x)})} + C \end{align} }