5.5 The Substitution Rule/35: Difference between revisions
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<math>\begin{align} | <math>\begin{align} | ||
let \; u &=1 + cos^2x \\[2ex] | let \; u &=1 + cos^2x dx \\[2ex] | ||
du &= 2cosx \;\cdot (-sinx) \\[2ex] | du &= 2cosx \;\cdot (-sinx) \\[2ex] | ||
-du & = 2sin(x)cos(x)\;dx & \\[2ex] | -du & = 2sin(x)cos(x)\;dx & \\[2ex] | ||
Revision as of 05:28, 5 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} \ \int \frac{sin 2 x}{1 + cos^2x} dx \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} let \; u &=1 + cos^2x dx \\[2ex] du &= 2cosx \;\cdot (-sinx) \\[2ex] -du & = 2sin(x)cos(x)\;dx & \\[2ex] -du & = sin(2x)dx & \end{align} }
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\ \int {\frac {sin2x}{1+cos^{2}x}}dx=-\ \int {\frac {du}{u}}=\ \int -ln(1+u)=\ \int |1+cos^{2}x|+c\end{aligned}}}