7.1 Integration By Parts/50: Difference between revisions

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&= sec^{n-2}(x) \cdot \tan(x) - (n-2)\int_{}^{} \left[\sec^{n-2}(x) \cdot \tan^{2}(x)\right]dx \\[2ex]
&= sec^{n-2}(x) \cdot \tan(x) - (n-2)\int_{}^{} \left[\sec^{n-2}(x) \cdot \tan^{2}(x)\right]dx \\[2ex]
&= sec^{n-2}(x) \cdot \tan(x) - (n-2)\int_{}^{} \left[\sec^{n-2}(x) \cdot \sec^{2}(x)-1]dx  \\[2ex]
&= sec^{n-2}(x) \cdot \tan(x) - (n-2)\int_{}^{} \left[\sec^{n-2}(x) \cdot [\sec^{2}(x)-1]dx  \\[2ex]
&= x \ln(x)^{n} - n \int_{}^{} \left(\ln(x)^{n-1} \right)dx \\[2ex]
&= x \ln(x)^{n} - n \int_{}^{} \left(\ln(x)^{n-1} \right)dx \\[2ex]


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

Revision as of 18:54, 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 (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}&u=\sec ^{n-2}(x)\quad dv=\sec ^{2}(x)dx\\[2ex]&du=(n-2)\sec ^{n-3}\cdot \sec(x)\tan(x)dx\quad v=\tan(x)\\[2ex]\end{aligned}}}

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \int_{}^{} \sec^{2}(x) \cdot \sec^{n-2}(x) dx &= \sec^{n-2}(x) \cdot \tan(x) - \int_{}^{} \left[(n-2)\sec^{n-3}(x) \cdot \sec(x)\tan(x)\right]\cdot \tan(x)dx \\[2ex] &= sec^{n-2}(x) \cdot \tan(x) - (n-2)\int_{}^{} \left[\sec^{n-2}(x) \cdot \tan^{2}(x)\right]dx \\[2ex] &= sec^{n-2}(x) \cdot \tan(x) - (n-2)\int_{}^{} \left[\sec^{n-2}(x) \cdot [\sec^{2}(x)-1]dx \\[2ex] &= x \ln(x)^{n} - n \int_{}^{} \left(\ln(x)^{n-1} \right)dx \\[2ex] \end{align} }