Question

integral de sen^2? heeelllpppp :)

Answer 1

$\displaystyle \int\sin^2(x)\,dx=\int\frac{1-\cos(2x)}{2}\,dx=\frac{1}{2}\int1-\cos(2x)\,dx\implies\\\\ \frac{1}{2}\int\,dx-\int\cos(2x)\,dx\implies 2x=u\implies dx=\frac{du}{2}\implies\\\\ \frac{1}{2}\int dx-\int\cos(u)\frac{du}{2}=\frac{1}{2}\int\,dx-\frac{1}{2}\int\cos\,(u)\,du\implies \\\\ \frac{1}{2}\left(x-\frac{1}{2}(\sin(u))\right)+C=\boxed{\frac{x}{2}-\frac{\sin(2x)}{4}+C}$