7.1 Integration By Parts/3: Difference between revisions
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<math> | <math> | ||
\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} | |||
</math> | |||
Latest revision as of 17:28, 16 December 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int \,x\cos 5xdx\\[2ex]u=x\qquad dv=\cos 5xdx\\[2ex]du=dx\qquad v={\frac {1}{5}}\sin 5x\\[2ex]\int \,x\cos 5xdx&=x\cdot {\frac {1}{5}}\sin 5x-\int {}^{}{\frac {1}{5}}\sin 5x\\[2ex]&=\int \,{\frac {1}{5}}\sin 5x\cdot \,dx\\[2ex]&={\frac {1}{5}}\int \,\sin 5xdx\\[2ex]&={\frac {1}{5}}\cdot {\frac {1}{5}}(-\cos 5x)\\[2ex]&=-{\frac {1}{25}}\cos 5x+C\\[2ex]&={\frac {1}{5}}x\sin 5x+{\frac {1}{25}}\cos 5x+C\end{aligned}}}