Arianac101603@students.laalliance.org at 05:23, 6 December 2022

2022-12-06T05:23:03Z

← Older revision		Revision as of 05:23, 6 December 2022
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	<math>\~~int_~~{0}^{1}\left(3+x\~~sqrt~~{x}\right)dx<math>		<math>\int\left(e^{2{\theta}}\right)sin{3\theta} d{\theta}</math>

			First:

			<math>u={\sin{(3\theta)}}</math>
			<math>du={3\cos{(3\theta)}} d{\theta}</math>

			<math>dv=\left(e^{2{\theta}}\right)</math>
			<math>v=\frac{1}{2}\left(e^{2{\theta}}\right)</math>

			Then:
			<math>I=\int\left(e^{2{\theta}}\right)\sin{3\theta} d{\theta}=\frac{1}{2}\left(e^{2{\theta}}\right)*{\sin{(3\theta)}}-\frac{3}{2}\int\left(e^{2{\theta}}\right)\cos{3\theta} d{\theta}</math>

			Next: let

			<math>U={\cos{(3\theta)}}</math>

			<math>dU={3\sin{(3\theta)}} d{\theta}</math>

			<math>dV=\left(e^{2{\theta}}\right) d{\theta}</math>

			<math>V=\frac{1}{2}\left(e^{2{\theta}}\right)</math> to get

			<math>\int\left(e^{2{\theta}}\right)\cos{3\theta} d{\theta}=\frac{1}{2}\int\left(e^{2{\theta}}\right)\cos{3\theta} d{\theta}+\frac{3}{2}\int\left(e^{2{\theta}}\right)\sin{3\theta} d{\theta}</math> substituting in the previous formula gives

			<math>I=\frac{1}{2}\left(e^{2{\theta}}\right)\sin{3\theta}-\frac{3}{4}\left(e^{2{\theta}}\right)\cos{3\theta}-\frac{9}{4}\int\left(e^{2{\theta}}\right)sin{3\theta} d{\theta}</math>

			<math>=\frac{1}{2}\left(e^{2{\theta}}\right)sin{3\theta}-\frac{3}{4}\left(e^{2{\theta}}\right)cos{3\theta}-\frac{9}{4}I</math>

			Then:
			<math>\frac{13}{4}I=\frac{1}{2}\left(e^{2{\theta}}\right)sin{3\theta}-\frac{3}{4}\left(e^{2{\theta}}\right)cos{3\theta}+C</math>,

			Hence,<math>I=\frac{1}{13}\left(e^{2{\theta}}\right)2sin{3\theta}-{3cos3\theta}+C</math>,

			Where
			<math>C=\frac{4}{13}C</math>

Arianac101603@students.laalliance.org: Created page with " $\int_{0}^{1}\left(3+x\sqrt{x}\right)dx "$

2022-12-06T04:02:52Z

Created page with "<math>\int_{0}^{1}\left(3+x\sqrt{x}\right)dx<math>"

New page

<math>\int_{0}^{1}\left(3+x\sqrt{x}\right)dx<math>

7.1 Integration By Parts/17 - Revision history

Arianac101603@students.laalliance.org at 05:23, 6 December 2022

Arianac101603@students.laalliance.org: Created page with "\int_{0}^{1}\left(3+x\sqrt{x}\right)dx"

Arianac101603@students.laalliance.org: Created page with " $\int_{0}^{1}\left(3+x\sqrt{x}\right)dx "$