5.5 The Substitution Rule/35

${\begin{aligned}\ \int {\frac {sin2x}{1+cos^{2}x}}dx\end{aligned}}$

${\begin{aligned}let\;u&=1+cos^{2}x\\[2ex]du&=2cosx\;\cdot (-sinx)dx\\[2ex]-du&=2sin(x)cos(x)\;dx&\\[2ex]-du&=sin(2x)dx&\end{aligned}}$

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} = \ \int -ln( 1 + u)= \ \int| 1 + cos^2x| + c \end{align} }