5.5 The Substitution Rule/31 - Revision history

Cynthiac61995@students.laalliance.org at 15:53, 4 October 2022

2022-10-04T15:53:06Z

← Older revision		Revision as of 15:53, 4 October 2022
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	\int \frac{\cos{(x)}}{\sin^2{(x)}}dx		\int \frac{\cos{(x)}}{\sin^2{(x)}}dx
	</math>		</math>


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


	<math>		<math>

Cynthiac61995@students.laalliance.org at 15:52, 4 October 2022

2022-10-04T15:52:42Z

← Older revision		Revision as of 15:52, 4 October 2022
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	\int \frac{\cos{(x)}}{\sin^2{(x)}}dx &= \int\frac{1}{u^2}du = \int u^{-2}du \\[2ex]		\int \frac{\cos{(x)}}{\sin^2{(x)}}dx &= \int\frac{1}{u^2}du = \int u^{-2}du \\[2ex]

	&= u^-1 + C \\[2ex]		&= u^{-1} + C \\[2ex]
	&= \frac{-1}{\sin{(x)}} + C		&= \frac{-1}{\sin{(x)}} + C

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

Cynthiac61995@students.laalliance.org at 15:52, 4 October 2022

2022-10-04T15:52:23Z

← Older revision		Revision as of 15:52, 4 October 2022
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	\begin{align}		\begin{align}

	\int \frac{\cos{(x)}}{\sin^2{(x)}}dx &= \int\frac{1}{u^2}du = \int u^(-2)du \\[2ex]		\int \frac{\cos{(x)}}{\sin^2{(x)}}dx &= \int\frac{1}{u^2}du = \int u^{-2}du \\[2ex]

	&= u^(-1) + C \\[2ex]		&= u^-1 + C \\[2ex]
	&= \frac{-1}{\sin{(x)}} + C		&= \frac{-1}{\sin{(x)}} + C

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

Cynthiac61995@students.laalliance.org at 15:51, 4 October 2022

2022-10-04T15:51:31Z

← Older revision		Revision as of 15:51, 4 October 2022
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	\begin{align}		\begin{align}

	\int \frac{\cos{(x)}}{\sin^2{(x)}}dx &= \int\frac{1}{u^2}du = \int u^-~~2du~~ \\[2ex]		\int \frac{\cos{(x)}}{\sin^2{(x)}}dx &= \int\frac{1}{u^2}du = \int u^(-2)du \\[2ex]

	&= u^-1 + C \\[2ex]		&= u^(-1) + C \\[2ex]
	&= \frac{-1}{\sin{(x)}} + C		&= \frac{-1}{\sin{(x)}} + C

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

Cynthiac61995@students.laalliance.org at 15:51, 4 October 2022

2022-10-04T15:51:01Z

← Older revision		Revision as of 15:51, 4 October 2022
Line 17:		Line 17:
	\int \frac{\cos{(x)}}{\sin^2{(x)}}dx &= \int\frac{1}{u^2}du = \int u^-2du \\[2ex]		\int \frac{\cos{(x)}}{\sin^2{(x)}}dx &= \int\frac{1}{u^2}du = \int u^-2du \\[2ex]

	&= u^-1 + C		&= u^-1 + C \\[2ex]
	&= \frac{-1}{\sin{(x)}} + C		&= \frac{-1}{\sin{(x)}} + C

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

Cynthiac61995@students.laalliance.org at 15:50, 4 October 2022

2022-10-04T15:50:29Z

← Older revision		Revision as of 15:50, 4 October 2022
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	u &=\sin{(x)} \\[2ex]		u &=\sin{(x)} \\[2ex]
	du &= \cos{(x)}dx		du &= \cos{(x)}dx

			\end{align}
			</math>

			<math>
			\begin{align}

			\int \frac{\cos{(x)}}{\sin^2{(x)}}dx &= \int\frac{1}{u^2}du = \int u^-2du \\[2ex]

			&= u^-1 + C
			&= \frac{-1}{\sin{(x)}} + C

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

Cynthiac61995@students.laalliance.org at 15:46, 4 October 2022

2022-10-04T15:46:12Z

← Older revision		Revision as of 15:46, 4 October 2022
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	<math>		<math>
	\int \frac{\cos{(x)}}{\sin^2{(x)}dx		\int \frac{\cos{(x)}}{\sin^2{(x)}}dx
	</math>		</math>

Cynthiac61995@students.laalliance.org: Created page with " $\int \frac{\cos{(x)}}{\sin^2{(x)}dx$ $\begin{align} u &=\sin{(x)} \\[2ex] du &= \cos{(x)}dx \end{align}$ "

2022-10-04T15:45:36Z

Created page with "<math> \int \frac{\cos{(x)}}{\sin^2{(x)}dx </math> <math> \begin{align} u &=\sin{(x)} \\[2ex] du &= \cos{(x)}dx \end{align} </math>"

New page

<math>
\int \frac{\cos{(x)}}{\sin^2{(x)}dx
</math>

<math>
\begin{align}

u &=\sin{(x)} \\[2ex]
du &= \cos{(x)}dx

\end{align}
</math>

5.5 The Substitution Rule/31 - Revision history

Cynthiac61995@students.laalliance.org at 15:53, 4 October 2022

Cynthiac61995@students.laalliance.org at 15:52, 4 October 2022

Cynthiac61995@students.laalliance.org at 15:52, 4 October 2022

Cynthiac61995@students.laalliance.org at 15:51, 4 October 2022

Cynthiac61995@students.laalliance.org at 15:51, 4 October 2022

Cynthiac61995@students.laalliance.org at 15:50, 4 October 2022

Cynthiac61995@students.laalliance.org at 15:46, 4 October 2022

Cynthiac61995@students.laalliance.org: Created page with " \int \frac{\cos{(x)}}{\sin^2{(x)}dx \begin{align} u &=\sin{(x)} \\[2ex] du &= \cos{(x)}dx \end{align} "

Cynthiac61995@students.laalliance.org: Created page with " $\int \frac{\cos{(x)}}{\sin^2{(x)}dx$ $\begin{align} u &=\sin{(x)} \\[2ex] du &= \cos{(x)}dx \end{align}$ "