5.5 The Substitution Rule/37: Difference between revisions
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u &= \sin(x) \\[2ex] | u &= \sin(x) \\[2ex] | ||
du &= | du &= \cos(x)dx \\[2ex] | ||
\frac{1}{3}du &= (a+bx^2)dx \\[2ex] | \frac{1}{3}du &= (a+bx^2)dx \\[2ex] | ||
Revision as of 18:56, 20 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 \begin{align} u &= \sin(x) \\[2ex] du &= \cos(x)dx \\[2ex] \frac{1}{3}du &= (a+bx^2)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{a+bx^2}{\sqrt{3ax+bx^3}}dx &= \int \frac{1}{\sqrt{3ax+bx^3}}(a+bx^2)\;dx = \int \frac{1}{\sqrt{3ax+bx^3}}(a+bx^2\;dx)\ \\[2ex] &= \frac{1}{3}\int \frac{1}{\sqrt{u}}(du) = \frac{1}{3}\int u^{-1/2} du \\[2ex] &= \frac{1}{3}\frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C \\[2ex] &= \frac{2}{3}(3ax+bx^3)^{1/2} + C \\[2ex] &= \frac{2}{3}{\sqrt{3ax+bx^3}} + C \end{align} }