5.5 The Substitution Rule/35

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

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

${\begin{aligned}\ \int {\frac {sin2x}{1+cos^{2}x}}dx=-\ \int {\frac {du}{u}}=\ \int -ln(1+u)=\ \int |1+cos^{2}x|+c\end{aligned}}$