7.1 Integration By Parts/3 - Revision history

Vanessav73764@students.laalliance.org at 17:28, 16 December 2022

2022-12-16T17:28:16Z

← Older revision		Revision as of 17:28, 16 December 2022
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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]

2022-12-16T10:47:15Z

← Older revision		Revision as of 10:47, 16 December 2022
Line 4:		Line 4:
	\int\,x\cos5xdx \\[2ex]		\int\,x\cos5xdx \\[2ex]
	u=x\qquad dv=\cos5xdx \\[2ex]		u=x\qquad dv=\cos5xdx \\[2ex]
	du=dx\qquad v=\frac{1}{5}\sin5x		du=dx\qquad v=\frac{1}{5}\sin5x\\[2ex]

			\int\,x\cos5xdx &=x\cdot\frac{1}{5}\sin5x-\int\\[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}		\end{align}

	</math>		</math>

2022-12-16T09:59:59Z

2022-12-15T19:55:30Z

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