∫ s i n ( π t ) d t {\displaystyle {\begin{aligned}\ \int sin(\pi t)dt\end{aligned}}}
l e t u = π t d u = π d x 1 π d u = d x {\displaystyle {\begin{aligned}let\;u&=\pi t\\[2ex]du&=\pi \;dx\\[2ex]{\frac {1}{\pi }}du&=dx\end{aligned}}}
∫ s i n ( π t ) d t = ∫ s i n ( u ) ( 1 π d u ) = ∫ 1 π ( − c o s u ) + c = ∫ − 1 π c o s ( π t ) + c {\displaystyle {\begin{aligned}\ \int sin(\pi t)dt=\int sin(u)({\frac {1}{\pi }}du)=\int {\frac {1}{\pi }}(-cosu)+c=\int -{\frac {1}{\pi }}cos(\pi t)+c\end{aligned}}}
![{\displaystyle {\begin{aligned}let\;u&=\pi t\\[2ex]du&=\pi \;dx\\[2ex]{\frac {1}{\pi }}du&=dx\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/49aac57364600374221d5071005045cce1063b7f)
