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Question Number 144527 by mnjuly1970 last updated on 26/Jun/21

$$$$$$.....\mathrm{Calculus}\left(\mathrm{I}\right).....$$$$\mathrm{P}:=\frac{{\int }_0^{\frac{\pi }{2}}(xcos\left(x\right)+1)e^{sin\left(x\right)}dx}{{\int }_0^{\frac{\pi }{2}}(xsin\left(x\right)-1)e^{cos\left(x\right)}dx}=?$$

Answered by Kamel last updated on 26/Jun/21

$$=\frac{{\left[{xe}^{sin\left(x\right)}\right]}_0^{\frac{\pi }{2}}}{-{\left[{xe}^{cos\left(x\right)}\right]}_0^{\frac{\pi }{2}}}=-e$$

Answered by Dwaipayan Shikari last updated on 26/Jun/21

$$\frac{d}{dx}\left({xe}^{f\left(x\right)}\right)={xe}^{f\left(x\right)}{f}^{\prime }\left(x\right)+e^{f\left(x\right)}$$$${xe}^{f\left(x\right)}=\int {xe}^{f\left(x\right)}{f}^{\prime }\left(x\right)+e^{f\left(x\right)}dx$$$$Heref\left(x\right)=sinxg\left(x\right)=cosx$$$$P=-\frac{{\left[{xe}^{sinx}\right]}_0^{\frac{\pi }{2}}}{{\left[{xe}^{cosx}\right]}_0^{\frac{\pi }{2}}}=-\frac{\frac{\pi e}{2}}{\frac{\pi }{2}}=-e$$

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