Answer to Question #110062 in Calculus for Nimra

Question #110062

∫²∫1(cotx + sinx)dx


1
Expert's answer
2020-04-30T18:48:19-0400

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\\

12cos(x)sint(x)dx=t=sin(x)dt=dsin(x)=cos(x)dx=sin(1)sin(2)dtt=lntsin(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))

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)


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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