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]


\text{Note: } \int \frac{1}{{x}}dx= ln(x)+C \\[2ex]
\text{Note: } \int \frac{1}{{x}}\;dx= ln(x)+C \\[2ex]


&= \left| ln(u) \right| + C \\[2ex]
&= \left| ln(u) \right| + C \\[2ex]

Revision as of 19:10, 20 September 2022



Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\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]{\text{Note: }}\int {\frac {1}{x}}\;dx=ln(x)+C\\[2ex]&=\left|ln(u)\right|+C\\[2ex]&=\left|ln(\sin(x))\right|+C\\[2ex]\end{aligned}}}