Mr. V Wiki Math - User contributions [en]

7.1 Integration By Parts/13

2022-12-16T05:40:03Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad \int\sec^2\left(2t\right)dt </math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

<math>
\int t\sec^2\left(2t\right) dt = \frac{1}{2}\tan\left(2t\right)-\frac{1}{2}\int\tan\left(2t\right)dt = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\int\tan\left(u\right)du = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\ln|\sec2t|+c
</math>

<math>
\begin{align}

& u=2t \\[2ex]
& du=2dt \\[2ex]
& \frac{du}{2}=dt \\[2ex]

\end{align}
</math>

7.1 Integration By Parts/13

2022-12-16T05:38:57Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad \int\sec^2\left(2t\right)dt </math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

<math>
= \frac{1}{2}\tan\left(2t\right)-\frac{1}{2}\int\tan\left(2t\right)dt = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\int\tan\left(u\right)du = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\ln|\sec2t|+c
</math>

<math>
\begin{align}

& u=2t \\[2ex]
& du=2dt \\[2ex]
& \frac{du}{2}=dt \\[2ex]

\end{align}
</math>

7.1 Integration By Parts/13

2022-12-16T05:38:37Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right)</math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

<math>
= \frac{1}{2}\tan\left(2t\right)-\frac{1}{2}\int\tan\left(2t\right)dt = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\int\tan\left(u\right)du = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\ln|\sec2t|+c
</math>

<math>
\begin{align}

& u=2t \\[2ex]
& du=2dt \\[2ex]
& \frac{du}{2}=dt \\[2ex]

\end{align}
</math>

7.1 Integration By Parts/13

2022-12-16T05:35:35Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad \int\sec^2\left(2t\right)dt </math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

<math>
= \frac{1}{2}\tan\left(2t\right)-\frac{1}{2}\int\tan\left(2t\right)dt = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\int\tan\left(u\right)du = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\ln|\sec2t|+c
</math>

<math>
\begin{align}

& u=2t \\[2ex]
& du=2dt \\[2ex]
& \frac{du}{2}=dt \\[2ex]

\end{align}
</math>

7.1 Integration By Parts/13

2022-12-16T05:34:37Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad \int\sec^2\left(2t\right)dt </math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

<math>
= \frac{1}{2}\tan\left(2t\right)-\frac{1}{2}\int\tan\left(2t\right)dt = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\int\tan\left(u\right)du = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\ln|\sec2t|+c

<math>
\begin{align}

& u=2t \\[2ex]
& du=2dt \\[2ex]
& \frac{du}{2}=dt \\[2ex]

\end{align}
</math>

7.1 Integration By Parts/13

2022-12-16T05:34:01Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad \int\sec^2\left(2t\right)dt </math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

<math>
= \frac{1}{2}\tan\left(2t\right)-\frac{1}{2}\int\tan\left(2t\right)dt = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\int\tan\left(u\right)du = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\ln|\sec2t|+c
<math>
\begin{align}

& u=2t
& du=2dt
& \frac{du}{2}=dt

\end{align}
</math>

7.1 Integration By Parts/13

2022-11-29T03:04:49Z

Andy Arevalo:

7.1 Integration By Parts/13

2022-11-29T03:04:40Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad &= \int\sec^2\left(2t\right)dt </math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

7.1 Integration By Parts/13

2022-11-29T03:03:55Z

Andy Arevalo:

7.1 Integration By Parts/13

2022-11-29T03:02:56Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad & = \int\sec^2\left(2t\right)dt </math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

7.1 Integration By Parts/13

2022-11-29T03:02:40Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad & = \int\sec^2\left(2t\right)dt </math>
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

7.1 Integration By Parts/13

2022-11-29T03:02:29Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad & = \int\sec^2\left(2t\right)dt \\[2ex] </math>
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

7.1 Integration By Parts/13

2022-11-29T03:02:14Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad & = \int\sec^2\left(2t\right)dt \\[2ex]
&= \frac{1}{2}\int\sec^2\left(u\right)du \\[2ex] </math>
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

7.1 Integration By Parts/13

2022-11-29T03:01:32Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad & = \int\sec^2\left(2t\right)dt \\[2ex]</math>
&= \frac{1}{2}\int\sec^2\left(u\right)du \\[2ex] </math>
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

7.1 Integration By Parts/13

2022-11-29T03:01:13Z

Andy Arevalo:

7.1 Integration By Parts/13

2022-11-29T03:01:03Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad & = \int\sec^2\left(2t\right)dt \\[2ex]
&= \frac{1}{2}\int\sec^2\left(u\right)du \\[2ex] </math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

7.1 Integration By Parts/13

2022-11-29T03:00:54Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad & = \int\sec^2\left(2t\right)dt \\[2ex]
&= \frac{1}{2}\int\sec^2\left(u\right)du \\[2ex]
</math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

7.1 Integration By Parts/13

2022-11-29T03:00:14Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad & = \int\sec^2\left(2t\right)dt
&= \frac{1}{2}\int\sec^2\left(u\right)du
</math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

7.1 Integration By Parts/13

2022-11-29T02:58:18Z

Andy Arevalo:

7.1 Integration By Parts/13

2022-11-29T02:57:58Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad \int\sec^2\left(2t\right)</math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

7.1 Integration By Parts/13

2022-11-29T02:56:09Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right)</math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

7.1 Integration By Parts/13

2022-11-29T02:56:00Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right)\quad dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right)</math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

7.1 Integration By Parts/13

2022-11-29T02:55:52Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right)\quaddt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right)</math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

7.1 Integration By Parts/13

2022-11-29T02:52:11Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math>u = t \qquad dv = \sec^2\left(2t\right)</math> 
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math>

7.1 Integration By Parts/13

2022-11-29T02:49:25Z

Andy Arevalo:

<math>\int t\sec^2\left(2t\right) dt </math>

<math> u = t \qquad dv = \sec^2\left(2t\right)

</math>

7.1 Integration By Parts/13

2022-11-29T02:48:51Z

Andy Arevalo:

<math>
\int t\sec^2\left(2t\right) dt
</math>

<math> u = t \qquad dv = \sec^2\left(2t\right)

</math>

7.1 Integration By Parts/13

2022-11-29T02:48:30Z

Andy Arevalo:

<math>
\int t\sec^2 2t dt
</math>

<math> u = t \qquad dv = \sec^2\left(2t\right)

</math>

7.1 Integration By Parts/13

2022-11-29T02:44:35Z

Andy Arevalo:

7.1 Integration By Parts/13

2022-11-29T02:43:51Z

Andy Arevalo:

7.1 Integration By Parts/13

2022-11-29T02:42:13Z

Andy Arevalo:

7.1 Integration By Parts/13

2022-11-29T02:41:59Z

Andy Arevalo:

7.1 Integration By Parts/13

2022-11-29T02:41:38Z

Andy Arevalo:

7.1 Integration By Parts/13

2022-11-29T02:41:29Z

Andy Arevalo: Created page with "<math> \inttsec^2 2tdt </math> <math> </math>"

<math>
\inttsec^2 2tdt
</math>

<math>

</math>

6.2 Volumes/3

2022-11-29T02:28:36Z

Andy Arevalo:

<math>
y=\frac{1}{x},
x = 1, x = 2, y = 0; \text{about the x-axis}
</math>

<math>
\begin{align}
\pi\int_1^2\left[\left(\frac{1}{x}\right)^2\right]dx & = \pi\int_1^2\left[\left(\frac{1}{x^2}\right)\right]dx \\[2ex]

&= \pi\left[-\frac{1}{x}\right]\Bigg|_1^2\\[2ex]
&= \pi\left[\left(-\frac{1}{\left(2\right)}\right)-\left(-\frac{1}{\left(1\right)}\right)\right] \\[2ex]
&= \pi\left[-\frac{1}{2}+1\right]= \pi\left[\frac{1}{2}\right] \\[2ex]
&= \frac{\pi}{2} \\[2ex]

\end{align}
</math>

6.2 Volumes/3

2022-11-29T02:28:19Z

Andy Arevalo:

<math>
y=\frac{1}{x},
x = 1, x = 2, y = 0; \text{about the x-axis}
</math>

<math>
\begin{align}
\pi\int_1^2\left[\left(\frac{1}{x}\right)^2\right]dx & = \pi\int_1^2\left[\left(\frac{1}{x^2}\right)\right]dx \\[2ex]

&= \pi\left[-\frac{1}{x}\right]\Bigg|_1^2\\[2ex]
&= \pi\left[\left(-\frac{1}{\left(2\right)}\right)-\left(-\frac{1}{\left(1\right)}\right)\right] \\[2ex]
&= \pi\left[-\frac{1}{2}+1\right= \pi\left[\frac{1}{2}\right] \\[2ex]
&= \frac{\pi}{2} \\[2ex]

\end{align}
</math>

6.2 Volumes/3

2022-11-29T02:27:51Z

Andy Arevalo:

<math>
y=\frac{1}{x},
x = 1, x = 2, y = 0; \text{about the x-axis}
</math>

<math>
\begin{align}
\pi\int_1^2\left[\left(\frac{1}{x}\right)^2\right]dx & = \pi\int_1^2\left[\left(\frac{1}{x^2}\right)\right]dx \\[2ex]

&= \pi\left[-\frac{1}{x}\right]\Bigg|_1^2\\[2ex]
&= \pi\left[\left(-\frac{1}{\left(2\right)}\right)-\left(-\frac{1}{\left(1\right)}\right)\right] \\[2ex]
&= \pi\left[-\frac{1}{2}+1\right]\Bigg|_1^2\\[2ex]= \pi\left[\frac{1}{2}\right] \\[2ex]
&= \frac{\pi}{2} \\[2ex]

\end{align}
</math>

6.2 Volumes/3

2022-11-29T02:27:23Z

Andy Arevalo:

<math>
y=\frac{1}{x},
x = 1, x = 2, y = 0; \text{about the x-axis}
</math>

<math>
\begin{align}
\pi\int_1^2\left[\left(\frac{1}{x}\right)^2\right]dx & = \pi\int_1^2\left[\left(\frac{1}{x^2}\right)\right]dx \\[2ex]

&= \pi\left[-\frac{1}{x}\right]\Bigg|_1^2\\[2ex]
&= \pi\left[\left(-\frac{1}{\left(2\right)}\right)-\left(-\frac{1}{\left(1\right)}\right)\right]\Bigg|_1^2 \\[2ex]
&= \pi\left[-\frac{1}{2}+1\right]\Bigg|_1^2\\[2ex]= \pi\left[\frac{1}{2}\right] \\[2ex]
&= \frac{\pi}{2} \\[2ex]

\end{align}
</math>

6.2 Volumes/3

2022-11-29T02:27:02Z

Andy Arevalo:

<math>
y=\frac{1}{x},
x = 1, x = 2, y = 0; \text{about the x-axis}
</math>

<math>
\begin{align}
\pi\int_1^2\left[\left(\frac{1}{x}\right)^2\right]dx & = \pi\int_1^2\left[\left(\frac{1}{x^2}\right)\right]dx \\[2ex]

&= \pi\left[-\frac{1}{x}\right]\Bigg|_1^2\\[2ex]
&= \pi\left[\left(-\frac{1}{\left(2\right)}\right)-\left(-\frac{1}{\left(1\right)}\right)\right]\Bigg|_1^2\\[2ex]
&= \pi\left[-\frac{1}{2}+1\right]\Bigg|_1^2\\[2ex]= \pi\left[\frac{1}{2}\right]\\[2ex]
&= \frac{\pi}{2}\\[2ex]

\end{align}
</math>

6.2 Volumes/3

2022-11-29T02:26:38Z

Andy Arevalo:

<math>
y=\frac{1}{x},
x = 1, x = 2, y = 0; \text{about the x-axis}
</math>

<math>
\begin{align}
\pi\int_1^2\left[\left(\frac{1}{x}\right)^2\right]dx & = \pi\int_1^2\left[\left(\frac{1}{x^2}\right)\right]dx \\[2ex]

&= \pi\left[-\frac{1}{x}\right]\Bigg|_1^2\\[2ex]
&= \pi\left[\left(-\frac{1}{\left(2\right)}\right)-\left(-\frac{1}{\left(1\right)}\right)\right]\Bigg|_1^2\\[2ex]
&= \pi\left[-\frac{1}{2}+1\right]\Bigg|_1^2\\[2ex]= \pi\left[\frac{1}{2}\right]
&= \frac{\pi}{2}

\end{align}
</math>

6.2 Volumes/3

2022-11-29T02:14:08Z

Andy Arevalo:

<math>
y=\frac{1}{x},
x = 1, x = 2, y = 0; \text{about the x-axis}
</math>

<math>
\begin{align}
\pi\int_1^2\left[\left(\frac{1}{x}\right)^2\right]dx & = \pi\int_1^2\left[\left(\frac{1}{x^2}\right)\right]dx \\[2ex]

&= \pi\left[-\frac{1}{x}\right]\Bigg|_1^2\\[2ex]
&= \pi\left[\left(-\frac{1}{\left(2\right)}\right)-\left(-\frac{1}{\left(1\right)}\right)\right]\Bigg|_1^2\\[2ex]
&= \pi\left[-\frac{1}{2}+1\right]\Bigg|_1^2\\[2ex]

\end{align}
</math>

6.2 Volumes/3

2022-11-29T02:12:32Z

Andy Arevalo:

6.2 Volumes/3

2022-11-29T02:12:05Z

Andy Arevalo:

<math>
y=\frac{1}{x},
x = 1, x = 2, y = 0; \text{about the x-axis}
</math>

<math>
\begin{align}
\pi\int_1^2\left[\left(\frac{1}{x}\right)^2\right]dx & = \pi\int_1^2\left[\left(\frac{1}{x^2}\right)\right]dx \\[2ex]

&= \pi\left[-\frac{1}{x}\right]\Bigg|_1^2\\[2ex]
&= \pi\left[\left(-\frac{1}{\left(2\right)}\right)-\left(-\frac{1}{1}\right)\right]\Bigg|_1^2\\[2ex]

\end{align}
</math>

6.2 Volumes/3

2022-11-29T02:10:37Z

Andy Arevalo:

<math>
y=\frac{1}{x},
x = 1, x = 2, y = 0; \text{about the x-axis}
</math>

<math>
\begin{align}
\pi\int_1^2\left[\left(\frac{1}{x}\right)^2\right]dx & = \pi\int_1^2\left[\left(\frac{1}{x^2}\right)\right]dx \\[2ex]

&= \pi\left[-\frac{1}{x}\right]\Bigg|_1^2\\[2ex]
&= \pi\left[\left(-\frac{1}{2}\right)-\left(-\frac{1}{1}\right)\right]\Bigg|_1^2\\[2ex]

\end{align}
</math>

6.2 Volumes/3

2022-11-29T02:08:57Z

Andy Arevalo:

<math>
y=\frac{1}{x},
x = 1, x = 2, y = 0; \text{about the x-axis}
</math>

<math>
\begin{align}
\pi\int_1^2\left[\left(\frac{1}{x}\right)^2\right]dx & = \pi\int_1^2\left[\left(\frac{1}{x^2}\right)\right]dx \\[2ex]

&= \pi\left[-\frac{1}{x}\right]\Bigg|_1^2\\[2ex]
&= \pi\left[left(-\frac{1}{2}\right)-\left(-\frac{1}{1}\right)\right]\Bigg|_1^2\\[2ex]

\end{align}
</math>

6.2 Volumes/3

2022-11-29T02:06:08Z

Andy Arevalo:

6.2 Volumes/3

2022-11-29T02:05:57Z

Andy Arevalo:

6.2 Volumes/3

2022-11-29T02:04:50Z

Andy Arevalo:

6.2 Volumes/3

2022-11-29T02:04:18Z

Andy Arevalo:

6.2 Volumes/3

2022-11-29T01:59:59Z

Andy Arevalo:

6.2 Volumes/3

2022-11-29T01:54:57Z

Andy Arevalo:

<math>
y=\frac{1}{x},
x = 1, x = 2, y = 0; \text{about the x-axis}
</math>

<math>
\begin{align}
\pi\int_1^2\left[\left(\frac{1}{x}\right)^2\right]dx

\end{align}
</math>