5.5 The Substitution Rule/54: Difference between revisions

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&= \int_{0}^{\pi} (\frac{1}{2}du)\cos{(u)} = \frac{1}{2}\int_{0}^{\pi} \cos{(u)}du \\[2ex]
&= \int_{0}^{\pi} (\frac{1}{2}du)\cos{(u)} = \frac{1}{2}\int_{0}^{\pi} \cos{(u)}du \\[2ex]
&= \frac{1}{2}\sin{(u)}\bigg|_{0}^{\pi} \\[2ex]
&= \frac{1}{2}\sin{(u)}\bigg|_{0}^{\pi} \\[2ex]
&= \frac{1}{2}\sin{((\pi)} - \frac{1}{2}\sin{((0)} \\[2ex]
&= \frac{1}{2}\sin{((\pi))} - \frac{1}{2}\sin{((0))} \\[2ex]


\end{align}
\end{align}
</math>
</math>

Revision as of 19:27, 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 (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{0}^{\sqrt {\pi }}x\cos {(x^{2})}\,dx&=\int _{0}^{\sqrt {\pi }}(xdx)\cos {(x^{2})}\\[2ex]&=\int _{0}^{\pi }({\frac {1}{2}}du)\cos {(u)}={\frac {1}{2}}\int _{0}^{\pi }\cos {(u)}du\\[2ex]&={\frac {1}{2}}\sin {(u)}{\bigg |}_{0}^{\pi }\\[2ex]&={\frac {1}{2}}\sin {((\pi ))}-{\frac {1}{2}}\sin {((0))}\\[2ex]\end{aligned}}}