∫²∫1(cotx + sinx)dx
I=∫12(cot(x)+sin(x))dx=∫12cot(x)dx+∫12sin(x)dxI=\int\limits_1^2(cot(x)+sin(x))dx=\int\limits_1^2cot(x)dx+\int\limits_1^2sin(x)dx\\I=1∫2(cot(x)+sin(x))dx=1∫2cot(x)dx+1∫2sin(x)dx
∫12cos(x)sint(x)dx=∣t=sin(x)dt=dsin(x)=cos(x)dx∣=∫sin(1)sin(2)dtt=ln∣t∣∣sin(1)sin(2)=ln(sin(2))−ln(sin(1))\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))1∫2sint(x)cos(x)dx=∣∣t=sin(x)dt=dsin(x)=cos(x)dx∣∣=sin(1)∫sin(2)tdt=ln∣t∣∣∣sin(1)sin(2)=ln(sin(2))−ln(sin(1))
∫12sin(x)dx=−cos(x)∣12=−cos(2)+cos(1)\int\limits_1^2sin(x)dx=-cos(x)\bigg|_1^2=-cos(2)+cos(1)1∫2sin(x)dx=−cos(x)∣∣12=−cos(2)+cos(1)
I=ln(sin(2))−ln(sin(1))−cos(2)+cos(1)I=ln(sin(2))-ln(sin(1))-cos(2)+cos(1)I=ln(sin(2))−ln(sin(1))−cos(2)+cos(1)
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