7.1 Integration By Parts/11: Difference between revisions

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(Created page with "<math> \int(arctan (4t)) dt </math> <math> \begin{align} \pi\int_1^2\left[\left(2-\frac{1}{2}x\right)^2\right]dx & = \pi\int_1^2\left[\left(4-2x+\frac{1}{4}x^2\right)\right]dx \\[2ex] &= \pi\left[4x-x^2+\frac{1}{12}x^3\right]\Bigg|_1^2 \\[2ex] &= \pi\left[\left(4(2)-(2)^2+\frac{1}{12}(2)^3\right)-\left(4(1)-(1)^2+\frac{1}{12}(1)^3\right)\right] \\[2ex] &= \pi\left[\left(8-4+\frac{8}{12}\right)-\left(4-1+\frac{1}{12}\right)\right] \\[2ex] &= \pi\left[4+\frac{8}{12}-3-\f...")
 
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<math>
<math>
\int(arctan (4t)) dt
\int \text {arctan(4t)}dt
</math>
</math>


<math>
<math>
\begin{align}
u= \text {arctan(4t)} \qquad dv=dt
\pi\int_1^2\left[\left(2-\frac{1}{2}x\right)^2\right]dx & = \pi\int_1^2\left[\left(4-2x+\frac{1}{4}x^2\right)\right]dx \\[2ex]
</math> <br><br>
 
<math>
du= \frac{4}{1+(4t)^{2}} dt \qquad v=t
</math>


&= \pi\left[4x-x^2+\frac{1}{12}x^3\right]\Bigg|_1^2 \\[2ex]
<math>
&= \pi\left[\left(4(2)-(2)^2+\frac{1}{12}(2)^3\right)-\left(4(1)-(1)^2+\frac{1}{12}(1)^3\right)\right] \\[2ex]
\int \text {arctan(4t)}dt = t \cdot \text {arctan(4t)}-4\int \frac{t}{1+16t^{2}} dt = t \cdot \text {arctan(4t)}-\frac{4}{32}\int\frac{1}{u} = t \cdot \text {arctan(4t)}-\frac{1}{8}\ln(u)= t \cdot \text {arctan(4t)}-\frac{1}{8}\ln(1+16t^{2})+C
&= \pi\left[\left(8-4+\frac{8}{12}\right)-\left(4-1+\frac{1}{12}\right)\right] \\[2ex]
</math>
&= \pi\left[4+\frac{8}{12}-3-\frac{1}{12}\right]= \pi\left[1+\frac{7}{12}\right] \\[2ex]
&= \pi\left[\frac{12}{12}+\frac{7}{12}\right]= \pi\left[\frac{19}{12}\right] \\[2ex]
&= \frac{19\pi}{12}


<math>
\begin{align}
& u=1+16t^{2} \\ [1ex]
& du=32t dt \\ [1ex]
& \frac{1}{32}du=t dt
\end{align}
\end{align}
</math>
</math>

Latest revision as of 05:55, 29 November 2022



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