Question Number 203573 by sonukgindia last updated on 22/Jan/24
Answered by esmaeil last updated on 22/Jan/24
![∫−e^(cosx) (1−cos^2 x−cosx)dx ∫−e^(cosx) (sin^2 x−cosx)dx= ∫−e^(cosx) sin^2 xdx+∫cosxe^(cosx) dx=Υ −e^(cosx) sinxdx=dv→ v=e^(cosx) sinx=u→cosxdx=du→ Υ=sinx(e^(cosx) )−∫e^(cosx) coscdx +∫e^(cosx) cosxdx=sinx(e^(cosx) )]_0 ^1 = sin(1)e^(cos1)](https://www.tinkutara.com/question/Q203585.png)
$$\int−{e}^{{cosx}} \left(\mathrm{1}−{cos}^{\mathrm{2}} {x}−{cosx}\right){dx} \\ $$$$\int−{e}^{{cosx}} \left({sin}^{\mathrm{2}} {x}−{cosx}\right){dx}= \\ $$$$\int−{e}^{{cosx}} {sin}^{\mathrm{2}} {xdx}+\int{cosxe}^{{cosx}} {dx}=\Upsilon \\ $$$$−{e}^{{cosx}} {sinxdx}={dv}\rightarrow \\ $$$${v}={e}^{{cosx}} \:\:\:{sinx}={u}\rightarrow{cosxdx}={du}\rightarrow \\ $$$$\Upsilon={sinx}\left({e}^{{cosx}} \right)−\int{e}^{{cosx}} {coscdx} \\ $$$$\left.+\int{e}^{{cosx}} {cosxdx}={sinx}\left({e}^{{cosx}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} = \\ $$$${sin}\left(\mathrm{1}\right){e}^{{cos}\mathrm{1}} \:\: \\ $$$$ \\ $$$$ \\ $$