Question

Алгебра пліз хелп мі

Answer 1

$I=\int_{0}^{\frac{\pi}{12}}\frac{dx}{sin^2\left ( x+\frac{\pi}{6} \right )}=\begin{vmatrix}t=x+\frac{\pi}{6}\\ dx=dt\end{vmatrix}=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{dt}{sin^2t}\\\left.\begin{matrix}I=-ctgt\end{matrix}\right|_{\frac{\pi}{6}}^{\frac{\pi}{4}}=\sqrt{3}-1$