7.1 Integration By Parts/43 - Revision history

Elizabethh58413 at 02:16, 29 November 2022

2022-11-29T02:16:00Z

@@ Line 8: / Line 8: @@
 &= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
 &= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]
 \end{align}
 </math>

2022-11-29T02:13:21Z

← Older revision		Revision as of 02:13, 29 November 2022
Line 1:		Line 1:
	<math> \text{Prove with the reduction formula that} \int\sin^2(x)dx = \frac{x}{2} - \frac{sin(2x)}{4} + C </math> <br>		<math> \text{Prove with the reduction formula that} \int\sin^2(x)dx = \frac{x}{2} - \frac{sin(2x)}{4} + C </math> <br>
	<math> \text{reduction formula:} \int\sin^ndx= -\frac{1}{n}\cos(x)\sin^{n-1}(x) + \frac{n-1}{n}\int\sin^{n-2}(x)dx </math>		<math> \text{reduction formula:} \int\sin^ndx= -\frac{1}{n}\cos(x)\sin^{n-1}(x) + \frac{n-1}{n}\int\sin^{n-2}(x)dx </math> <br>
	<math>		<math>
	\begin{align}		\begin{align}

2022-11-29T00:26:18Z

← Older revision		Revision as of 00:26, 29 November 2022
Line 1:		Line 1:
	<math> \text{Prove with the reduction formula that} \int\sin^2(x)dx = \frac{x}{2} - \frac{sin(2x)}{4} + C </math~~> <br~~><br>		<math> \text{Prove with the reduction formula that} \int\sin^2(x)dx = \frac{x}{2} - \frac{sin(2x)}{4} + C </math> <br>
	<math> \text{reduction formula:} \int\sin^ndx= -\frac{1}{n}\cos(x)\sin^{n-1}(x) + \frac{n-1}{n}\int\sin^{n-2}(x)dx </math>		<math> \text{reduction formula:} \int\sin^ndx= -\frac{1}{n}\cos(x)\sin^{n-1}(x) + \frac{n-1}{n}\int\sin^{n-2}(x)dx </math>
	<math>		<math>

2022-11-29T00:25:05Z

← Older revision		Revision as of 00:25, 29 November 2022
Line 1:		Line 1:
	<math> \text{Prove with the reduction formula that} \int\sin^2(x)dx = \frac{x}{2} - \frac{sin(2x)}{4} + C </math>		<math> \text{Prove with the reduction formula that} \int\sin^2(x)dx = \frac{x}{2} - \frac{sin(2x)}{4} + C </math> <br><br>
	<math> \text{reduction formula:} \int\sin^ndx= -\frac{1}{n}\cos(x)\sin^{n-1}(x) + \frac{n-1}{n}\int\sin^{n-2}(x)dx </math>		<math> \text{reduction formula:} \int\sin^ndx= -\frac{1}{n}\cos(x)\sin^{n-1}(x) + \frac{n-1}{n}\int\sin^{n-2}(x)dx </math>
	<math>		<math>

2022-11-29T00:24:46Z

← Older revision		Revision as of 00:24, 29 November 2022
Line 1:		Line 1:
	<math>		<math> \text{Prove with the reduction formula that} \int\sin^2(x)dx = \frac{x}{2} - \frac{sin(2x)}{4} + C </math>
			<math> \text{reduction formula:} \int\sin^ndx= -\frac{1}{n}\cos(x)\sin^{n-1}(x) + \frac{n-1}{n}\int\sin^{n-2}(x)dx </math>
			<math>
	\begin{align}		\begin{align}

2022-11-29T00:23:44Z

← Older revision		Revision as of 00:23, 29 November 2022
Line 3:		Line 3:

	\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx \\[2ex]		\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx \\[2ex]

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

	~~\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx \\[2ex]~~


	&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]		&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
	&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]		&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
	&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]		&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]
			\end{align}
			</math>

2022-11-29T00:23:05Z

← Older revision		Revision as of 00:23, 29 November 2022
Line 1:		Line 1:
	<math>		<math>
	\begin{align}		\begin{align}
	\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x +
			\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx \\[2ex]

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

	\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx ~~</math>~~ \\[2ex]		\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx \\[2ex]


	&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]		&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
	&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]		&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
	&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]		&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]

2022-11-29T00:22:28Z

← Older revision		Revision as of 00:22, 29 November 2022
Line 1:		Line 1:
	<math>		<math>
	\begin{align}		\begin{align}
	\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot		\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x +

	\end{align}		\end{align}

2022-11-29T00:22:15Z

2022-11-29T00:21:55Z