7.1 Integration By Parts/45 - Revision history

Johnny C. at 00:30, 30 November 2022

2022-11-30T00:30:09Z

@@ Line 43: / Line 43: @@
@@ Line 91: / Line 61: @@
 </math>
 \begin{align}
@@ Line 98: / Line 68: @@
 &= \frac{4}{5}\int_{0}^{\frac{\pi}{2}} (1-cos^2(x)) sin(x) dx \\[2ex]
 &= -\int_{1}^{0} (1-u^2)\cdot du \\[2ex]
-&= \frac{4}{5} \left[-u + \frac{u^3}{3}\right]\bigg|_{0}^{1} \\[2ex]
+&= \frac{4}{5} \left[u - \frac{u^3}{3}\right]\bigg|_{0}^{1} \\[2ex]
@@ Line 105: / Line 75: @@
 &  dx= \frac{du}{-sin(x)} \\[2ex]
 & \int sin^3(x) dx \\[2ex]
-&= -\int 1-u^2 du \\[2ex]
+&= \frac{4}{5} [1-\frac{1}{3}] \\[2ex]
-&= \frac{4}{5} [-cos(x) + \frac{1}{3} cos^3(x)] \\[2ex]
+&= \frac{8}{15}
 \end{align}

Dvaezazizi@laalliance.org at 21:17, 29 November 2022

2022-11-29T21:17:14Z

← Older revision		Revision as of 21:17, 29 November 2022
Line 90:		Line 90:

	</math>		</math>


			----
			\begin{align}

			\int_{0}^{\frac{\pi}{2}} sin^5(x) dx &= \frac{5-1}{5} \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex]
			&= \frac{4}{5}\int_{0}^{\frac{\pi}{2}} (1-cos^2(x)) sin(x) dx \\[2ex]
			&= -\int_{1}^{0} (1-u^2)\cdot du \\[2ex]
			&= \frac{4}{5} \left[-u + \frac{u^3}{3}\right]\bigg\|_{0}^{1} \\[2ex]


			& u= cos(x) \\[2ex]
			& du= -sin(x) \\[2ex]
			& dx= \frac{du}{-sin(x)} \\[2ex]
			& \int sin^3(x) dx \\[2ex]
			&= -\int 1-u^2 du \\[2ex]
			&= \frac{4}{5} [-cos(x) + \frac{1}{3} cos^3(x)] \\[2ex]
			&= \frac{4}{5} [-cos(\frac{\pi}{2}) + \frac{1}{3} cos^3 (\frac{\pi}{2}) - (-cos(0) + \frac{1}{3} cos^3 (0)) \\[2ex]
			&= \frac{4}{5} [(-1)(0) + \frac{1}{3} (0)^3 - ((-1)(1) +\frac{1}{3} cos^3(1) \\[2ex]

			&= \frac{4}{5} [-(-1+ \frac{1}{3}(1)] \\[2ex]
			&= \frac{4}{5}[\frac{2}{3}] \\[2ex]
			&= \frac{8}{15} \\[2ex]

			\end{align}

Dvaezazizi@laalliance.org at 21:11, 29 November 2022

2022-11-29T21:11:11Z

← Older revision		Revision as of 21:11, 29 November 2022
Line 48:		Line 48:


	&\int_{0}^{\frac{\pi}{2}} sin^5(x) dx \\[2ex]		\int_{0}^{\frac{\pi}{2}} sin^5(x) dx \\[2ex]

	&= \frac{5-1}{5} \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex]		&= \frac{5-1}{5} \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex]

Johnny C. at 05:52, 29 November 2022

2022-11-29T05:52:02Z

← Older revision		Revision as of 05:52, 29 November 2022
Line 63:		Line 63:
	&= \frac{4}{5} [(-1)(0) + \frac{1}{3} (0)^3 - ((-1)(1) +\frac{1}{3} cos^3(1) \\[2ex]		&= \frac{4}{5} [(-1)(0) + \frac{1}{3} (0)^3 - ((-1)(1) +\frac{1}{3} cos^3(1) \\[2ex]

	&= \frac{4}{5} [-(-1+ \frac{1}{3}(1)]		&= \frac{4}{5} [-(-1+ \frac{1}{3}(1)] \\[2ex]
	&= \frac{4}{5}[\frac{2}{3}]		&= \frac{4}{5}[\frac{2}{3}] \\[2ex]
	&= \frac{8}{15}		&= \frac{8}{15} \\[2ex]





			\end{align}

			</math>


			<math>

			\begin{align}

			& \int_{0}^{\frac{\pi}{2}} sin^{2n+1} (x) dx \\[2ex]
			&= \frac{2n+1-1}{2n+1} \int_{0}^{\frac{\pi}{2}} sin^{2n+1-2} (x) dx \\[2ex]
			&= \frac{2n}{2n+1} \int_{0}^{\frac{\pi}{2}} sin^{2n-1} (x) dx \\[2ex]

Johnny C.: Created page with " $\begin{align} & \int_{0}^{\frac{\pi}{2}} sin^n(x) dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2} (x) dx \\[2ex] &= -\frac{1}{n} cos(x) sin^{n-1} (x) + \frac{n-1}{n} \bigg|_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex] &= -\frac{1}{n} cos(\frac{\pi}{2}) sin^{n-1} (\frac{\pi}{2})+ \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex] &= -\frac{1}{n} (0) (1) + \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex] &..."$

2022-11-29T05:48:48Z

Created page with "<math> \begin{align} & \int_{0}^{\frac{\pi}{2}} sin^n(x) dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2} (x) dx \\[2ex] &= -\frac{1}{n} cos(x) sin^{n-1} (x) + \frac{n-1}{n} \bigg|_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex] &= -\frac{1}{n} cos(\frac{\pi}{2}) sin^{n-1} (\frac{\pi}{2})+ \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex] &= -\frac{1}{n} (0) (1) + \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex] &..."

New page

<math>

\begin{align}

& \int_{0}^{\frac{\pi}{2}} sin^n(x) dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2} (x) dx \\[2ex]

&= -\frac{1}{n} cos(x) sin^{n-1} (x) + \frac{n-1}{n} \bigg|_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]
&= -\frac{1}{n} cos(\frac{\pi}{2}) sin^{n-1} (\frac{\pi}{2})+ \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]
&= -\frac{1}{n} (0) (1) + \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]
&= \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]

\end{align}

</math>

<math>

\begin{align}

& \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex]

&= \frac{3-1}{3} \int_{0}^{\frac{\pi}{2}} sin(x) dx \\[2ex]
&= \frac{2}{3} \int_{0}^{\frac{\pi}{2}} sin(x) dx \\[2ex]
&= \frac{2}{3} [-cos(x) \bigg|_{0}^{\frac{\pi}{2}} \\[2ex]
&= \frac{2}{3} [-cos(\frac{\pi}{2}) -(-cos(0))] \\[2ex]
&= \frac{2}{3} [-1(0) -(-1)] \\[2ex]
&= \frac{2}{3}[1] \\[2ex]
&= \frac{2}{3} \\[2ex]

\end{align}

</math>

<math>

\begin{align}

&\int_{0}^{\frac{\pi}{2}} sin^5(x) dx \\[2ex]

&= \frac{5-1}{5} \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex]
& u= cos(x) \\[2ex]
& du= -sin(x) \\[2ex]
& dx= \frac{du}{-sin(x)} \\[2ex]
& \int sin^3(x) dx \\[2ex]
&= \int (1-cos^2(x)) sin(x) dx \\[2ex]
&= \int (1-u^2)sin(x) \cdot \frac{du}{-sin(x)} \\[2ex]
&= -\int 1-u^2 du \\[2ex]
&= \frac{4}{5} [-u + \frac{1}{3}^3(x)] \\[2ex]
&= \frac{4}{5} [-cos(x) + \frac{1}{3} cos^3(x)] \\[2ex]
&= \frac{4}{5} [-cos(\frac{\pi}{2}) + \frac{1}{3} cos^3 (\frac{\pi}{2}) - (-cos(0) + \frac{1}{3} cos^3 (0)) \\[2ex]
&= \frac{4}{5} [(-1)(0) + \frac{1}{3} (0)^3 - ((-1)(1) +\frac{1}{3} cos^3(1) \\[2ex]

&= \frac{4}{5} [-(-1+ \frac{1}{3}(1)]
&= \frac{4}{5}[\frac{2}{3}]
&= \frac{8}{15}

\end{align}

</math>