∫ s i n 2 x 1 + c o s 2 x d x {\displaystyle {\begin{aligned}\ \int {\frac {sin2x}{1+cos^{2}x}}dx\end{aligned}}}
l e t u = 1 + c o s 2 x d u = 2 c o s x ⋅ ( − s i n x ) d x − d u = 2 s i n ( x ) c o s ( x ) d x − d u = s i n ( 2 x ) d x {\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}}}
∫ s i n 2 x 1 + c o s 2 x d x = − ∫ d u u = − l n ( 1 + u ) + c = − | 1 + c o s 2 x | + c {\displaystyle {\begin{aligned}\ \int {\frac {sin2x}{1+cos^{2}x}}dx\;=\;-\ \int {\frac {du}{u}}\;=\;-ln(1+u)+c\;=\;-|1+cos^{2}x|+c\end{aligned}}}
![{\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)
