7.1 Integration By Parts/50: Difference between revisions

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<math>
<math>
\begin{align}
\begin{align}
&u = \ln(x)^{n} \quad dv= 1dx \\[2ex]
&u = \sec^{n-2}(x) \quad dv= \sec^{2}(x)dx \\[2ex]
&du =1dx       \quad v=x \\[2ex]
&du = (n-2)\sec^{n-3} \cdot \sec(x) \cdot \tan(x) dx       \quad v= \tan(x) \\[2ex]


\end{align}
\end{align}

Revision as of 18:33, 29 November 2022

Prove 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_{}^{} \sec^{n}(x)dx = \frac{\tan(x) \cdot \sec^{n-2}(x)}{n-1} + \frac{n-2}{n-1} \int_{}^{} \sec^{n-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 \int_{}^{} \sec^{n}(x)dx = \int_{}^{} \sec^{2}(x) \cdot \sec^{n-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 = \sec^{n-2}(x) \quad dv= \sec^{2}(x)dx \\[2ex] &du = (n-2)\sec^{n-3} \cdot \sec(x) \cdot \tan(x) dx \quad v= \tan(x) \\[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_{}^{} \left(\ln(x)^{n}\right)dx &= x \ln(x)^{n} - \int_{}^{} \left((x \frac{n \ln(x)^{n-1}}{x}) \right)dx \\[2ex] &= x \ln(x)^{n} - \int_{}^{} \left(n \ln(x)^{n-1} \right)dx \\[2ex] &= x \ln(x)^{n} - n \int_{}^{} \left(\ln(x)^{n-1} \right)dx \\[2ex] \end{align} }