Question Number 144527 by mnjuly1970 last updated on 26/Jun/21
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:…..\:\:\:\mathrm{Calculus}\:\:\left(\mathrm{I}\:\right)….. \\ $$$$\mathrm{P}:=\:\frac{\int_{\mathrm{0}\:} ^{\:\:\frac{\pi}{\mathrm{2}}} \left(\:{xcos}\left({x}\right)+\mathrm{1}\:\right){e}^{\:{sin}\left({x}\right)} {dx}\:}{\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \left(\:{xsin}\left({x}\right)\:−\mathrm{1}\:\right){e}^{\:{cos}\left({x}\:\right)} {dx}}=? \\ $$
Answered by Kamel last updated on 26/Jun/21
![=(([xe^(sin(x)) ]_0 ^(π/2) )/(−[xe^(cos(x)) ]_0 ^(π/2) ))=−e](https://www.tinkutara.com/question/Q144543.png)
$$\:\:\:\:\:=\frac{\left[{xe}^{{sin}\left({x}\right)} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} }{−\left[{xe}^{{cos}\left({x}\right)} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} }=−{e} \\ $$
Answered by Dwaipayan Shikari last updated on 26/Jun/21
![(d/dx)(xe^(f(x)) )=xe^(f(x)) f′(x)+e^(f(x)) xe^(f(x)) =∫xe^(f(x)) f′(x)+e^(f(x)) dx Here f(x)=sinx g(x)=cosx P=−(([xe^(sinx) ]_0 ^(π/2) )/([xe^(cosx) ]_0 ^(π/2) ))=−(((πe)/2)/(π/2))=−e](https://www.tinkutara.com/question/Q144555.png)
$$\frac{{d}}{{dx}}\left({xe}^{{f}\left({x}\right)} \right)={xe}^{{f}\left({x}\right)} {f}'\left({x}\right)+{e}^{{f}\left({x}\right)} \\ $$$${xe}^{{f}\left({x}\right)} =\int{xe}^{{f}\left({x}\right)} {f}'\left({x}\right)+{e}^{{f}\left({x}\right)} {dx} \\ $$$${Here}\:{f}\left({x}\right)={sinx}\:\:\:\:{g}\left({x}\right)={cosx} \\ $$$${P}=−\frac{\left[{xe}^{{sinx}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} }{\left[{xe}^{{cosx}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} }=−\frac{\frac{\pi{e}}{\mathrm{2}}}{\frac{\pi}{\mathrm{2}}}=−{e} \\ $$