5.5 The Substitution Rule/37: Difference between revisions

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&= \int \frac{1}{{u}}(du) \\[2ex]
&= \int \frac{1}{{u}}(du) \\[2ex]
<math>\text{Note: } \int \frac{1}{{x}}dx=\ln(x)+C</math>
&= \left| \ln(u) \right| + C \\[2ex]
&= \frac{1}{3}\frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C \\[2ex]
&= \left| \ln(\sin(x)) \right| + C \\[2ex]
&= \frac{2}{3}(3ax+bx^3)^{1/2} + C \\[2ex]
 
&= \frac{2}{3}{\sqrt{3ax+bx^3}} + C


\end{align}
\end{align}
</math>
</math>

Latest revision as of 19:18, 20 September 2022


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} u &= \sin(x) \\[2ex] du &= \cos(x)\;dx \\[2ex] \end{align} }


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int \frac{\cos(x)}{\sin(x)}dx &= \int \frac{1}{\sin(x)}\cos(x)\;dx = \int \frac{1}{\sin(x)}(\cos(x)\;dx) \\[2ex] &= \int \frac{1}{{u}}(du) \\[2ex] &= \left| \ln(u) \right| + C \\[2ex] &= \left| \ln(\sin(x)) \right| + C \\[2ex] \end{align} }