7.1 Integration By Parts/3: Difference between revisions

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du=dx\qquad v=\frac{1}{5}\sin5x\\[2ex]
du=dx\qquad v=\frac{1}{5}\sin5x\\[2ex]


\int\,x\cos5xdx &=x\cdot\frac{1}{5}\sin5x-\int\\[2ex]
\int\,x\cos5xdx &=x\cdot\frac{1}{5}\sin5x-\int{}^{}\frac{1}{5}\sin5x\\[2ex]
 
&= \int\,\frac{1}{5}\sin5x\cdot\,dx\\[2ex]
&= \int\,\frac{1}{5}\sin5x\cdot\,dx\\[2ex]



Latest revision as of 17:28, 16 December 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} \int\,x\cos5xdx \\[2ex] u=x\qquad dv=\cos5xdx \\[2ex] du=dx\qquad v=\frac{1}{5}\sin5x\\[2ex] \int\,x\cos5xdx &=x\cdot\frac{1}{5}\sin5x-\int{}^{}\frac{1}{5}\sin5x\\[2ex] &= \int\,\frac{1}{5}\sin5x\cdot\,dx\\[2ex] &= \frac{1}{5}\int\,\sin5xdx\\[2ex] &= \frac{1}{5}\cdot\frac{1}{5}(-\cos5x)\\[2ex] &= -\frac{1}{25}\cos5x+C\\[2ex] &= \frac{1}{5}x\sin5x+\frac{1}{25}\cos5x+C \end{align} }