5.5 The Substitution Rule/30: Difference between revisions
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(Created page with "<math> \int \frac{\sin{\ln{(x)}}}{x}dx </math> <math> \begin{align} u &=2+x^4 \\[2ex] du &= 4x^3dx \\[2ex] \frac{1}{4}du &= x^3dx \end{align} </math> <math> \begin{align} \int x^3(2+x^4)^5dx &= \int (x^3dx)(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex] &= \frac{1}{4} \left[\frac{u^{1+1}}{1+1}\right] + C = \frac{u^2}{8} + C \\[2ex] &= \frac{(2+x^4)^2}{8} + C \end{align} </math>") |
m (Protected "5.5 The Substitution Rule/30" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))) |
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<math> | <math> | ||
\int \frac{\sin{\ln{(x)}}}{x}dx | \int \frac{\sin{(\ln{(x))}}}{x}dx | ||
</math> | </math> | ||
| Line 7: | Line 7: | ||
\begin{align} | \begin{align} | ||
u &= | u &=\ln(x) \\[2ex] | ||
du &= | du &= \frac{1}{x}dx \\[2ex] | ||
\frac{1}{ | |||
\end{align} | \end{align} | ||
| Line 17: | Line 16: | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
&= \frac{1}{ | \int \frac{\sin{(\ln{(x))}}}{x}dx &= \int\frac{1}{x}\sin(\ln{(x)})dx = \int\left(\frac{1}{x}dx\right)\sin{(\ln{(x)})} \\[2ex] | ||
&= \ | &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] | ||
&= -\cos{(u)} + C \\[2ex] | |||
&= -\cos{(\ln{(x)})} + C | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Latest revision as of 19:10, 26 August 2022
![{\displaystyle {\begin{aligned}u&=\ln(x)\\[2ex]du&={\frac {1}{x}}dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/2dcd1cacb47fe0ded9f0cb8411297ff136d61465)
![{\displaystyle {\begin{aligned}\int {\frac {\sin {(\ln {(x))}}}{x}}dx&=\int {\frac {1}{x}}\sin(\ln {(x)})dx=\int \left({\frac {1}{x}}dx\right)\sin {(\ln {(x)})}\\[2ex]&=\int (du)\sin {(u)}=\int \sin {(u)}du\\[2ex]&=-\cos {(u)}+C\\[2ex]&=-\cos {(\ln {(x)})}+C\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/92ec083cbb58cbf42efe622bfc22fd4c76bb337f)