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

Revision as of 19:05, 20 September 2022



Failed to parse (unknown function "\begin{align}"): {\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] <math>\text{Note: } \int \frac{1}{{x}}dx= ln(x)+C }

&= \frac{1}{3}\frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C \\[2ex] &= \frac{2}{3}(3ax+bx^3)^{1/2} + C \\[2ex] &= \frac{2}{3}{\sqrt{3ax+bx^3}} + C

\end{align} </math>