7.1 Integration By Parts/49 - Revision history

Ricardom59381@students.laalliance.org at 04:36, 30 November 2022

2022-11-30T04:36:00Z

2022-11-30T04:35:31Z

← Older revision		Revision as of 04:35, 30 November 2022
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	Prove		Prove
	<math>		<math>
	\int_{}^{} \left(\tan^{n}(x)\right)dx =\frac{\tan^{n-1}x}{n-1} - \int_{}^{} \left(\tan^{n-2}x\right)dx ~~\\[2ex]~~		\int_{}^{} \left(\tan^{n}(x)\right)dx =\frac{\tan^{n-1}x}{n-1} - \int_{}^{} \left(\tan^{n-2}x\right)dx

	</math>		</math>

2022-11-30T04:35:23Z

← Older revision		Revision as of 04:35, 30 November 2022
Line 1:		Line 1:
	Prove		Prove
	<math>		<math>
	\int_{}^{} \left(\tan^{n}(x)\right)dx =\frac{\tan^{n-1}x}{n-1} - \int_{}^{} \left(\tan^{n-2}x\right)dx		\int_{}^{} \left(\tan^{n}(x)\right)dx =\frac{\tan^{n-1}x}{n-1} - \int_{}^{} \left(\tan^{n-2}x\right)dx \\[2ex]

	</math> ~~\\[2ex]~~		</math>

	Note:		Note:

2022-11-30T04:35:12Z

← Older revision		Revision as of 04:35, 30 November 2022
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	\int_{}^{} \left(\tan^{n}(x)\right)dx =\frac{\tan^{n-1}x}{n-1} - \int_{}^{} \left(\tan^{n-2}x\right)dx		\int_{}^{} \left(\tan^{n}(x)\right)dx =\frac{\tan^{n-1}x}{n-1} - \int_{}^{} \left(\tan^{n-2}x\right)dx

	</math>		</math> \\[2ex]

	Note:		Note:

2022-11-30T04:34:42Z

@@ Line 3: / Line 3: @@
 \int_{}^{} \left(\tan^{n}(x)\right)dx =\frac{\tan^{n-1}x}{n-1} - \int_{}^{} \left(\tan^{n-2}x\right)dx
 </math>
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 = \frac{\tan^{n-1}(x)}{n-1} -\int_{}^{}\tan^{n-2}(x)dx
 \end{align}
 </math>

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← Older revision		Revision as of 04:30, 30 November 2022
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	\frac{\tan^{n-1}(x)}{n-1} = \frac{(n-1)}{n-1} \int_{}^{} (\sec^{2}x)(\tan^{n-2}x)dx		\frac{\tan^{n-1}(x)}{n-1} = \frac{(n-1)}{n-1} \int_{}^{} (\sec^{2}x)(\tan^{n-2}x)dx
	Bring down -\int_{}^{}\tan^{n-2}(x)dx
			\end{align}
			</math>

			Bring down

			</math>
			\begin{align}

			-\int_{}^{}\tan^{n-2}(x)dx
	= \frac{\tan^{n-1}(x)}{n-1} -\int_{}^{}\tan^{n-2}(x)dx		= \frac{\tan^{n-1}(x)}{n-1} -\int_{}^{}\tan^{n-2}(x)dx