5.5 The Substitution Rule/35: Difference between revisions
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<math>\begin{align} \ \int \frac{sin 2 x}{1 + cos^2x} dx = -\ \int \frac {du}{u} = -ln( 1 + u)= -| 1 + cos^2x| + c | <math>\begin{align} \ \int \frac{sin 2 x}{1 + cos^2x} dx = -\ \int \frac {du}{u} = -ln( 1 + u) = c = -| 1 + cos^2x| + c | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 05:30, 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} }
![{\displaystyle {\begin{aligned}let\;u&=1+cos^{2}x\\[2ex]du&=2cosx\;\cdot (-sinx)\;dx\\[2ex]-du&=2sin(x)cos(x)\;dx&\\[2ex]-du&=sin(2x)dx&\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/9ae18dc0dc98cbf51f41b3c7d59c91dffa33ceef)
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 = -\ \int \frac {du}{u} = -ln( 1 + u) = c = -| 1 + cos^2x| + c \end{align} }