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 | \text{Note: } \int \frac{1}{{x}}dx= ln(x)+C \\[2ex] | ||
&= \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] | ||
Revision as of 19:06, 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] \text{Note: } \int \frac{1}{{x}}dx= ln(x)+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}{\sqrt{3ax+bx^3}} + C \end{align} }