Answer to Question #110062 in Calculus for Nimra

Question

∫²∫¹(cotx + sinx)dx

Answer 1

$I=\int\limits_1^2(cot(x)+sin(x))dx=\int\limits_1^2cot(x)dx+\int\limits_1^2sin(x)dx\\$

$\int\limits_1^2\frac{cos(x)}{sint(x)}dx=\begin{vmatrix}t=sin(x)\\dt=dsin(x)=cos(x)dx\end{vmatrix}=\int\limits_{sin(1)}^{sin(2)}\frac{dt}{t}=\\ ln|t|\bigg|_{sin(1)}^{sin(2)}=ln(sin(2))-ln(sin(1))$

$\int\limits_1^2sin(x)dx=-cos(x)\bigg|_1^2=-cos(2)+cos(1)$

$I=ln(sin(2))-ln(sin(1))-cos(2)+cos(1)$