5.5 The Substitution Rule/31: Difference between revisions

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\begin{align}
\begin{align}


\int \frac{\cos{(x)}}{\sin^2{(x)}}dx &= \int\frac{1}{u^2}du = \int u^-2du \\[2ex]
\int \frac{\cos{(x)}}{\sin^2{(x)}}dx &= \int\frac{1}{u^2}du = \int u^(-2)du \\[2ex]


&= u^-1 + C \\[2ex]
&= u^(-1) + C \\[2ex]
&= \frac{-1}{\sin{(x)}} + C
&= \frac{-1}{\sin{(x)}} + C


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

Revision as of 15:51, 4 October 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 \frac{\cos{(x)}}{\sin^2{(x)}}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 &=\sin{(x)} \\[2ex] du &= \cos{(x)}dx \end{align} }

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int {\frac {\cos {(x)}}{\sin ^{2}{(x)}}}dx&=\int {\frac {1}{u^{2}}}du=\int u^{(}-2)du\\[2ex]&=u^{(}-1)+C\\[2ex]&={\frac {-1}{\sin {(x)}}}+C\end{aligned}}}