Question Number 65485 by aliesam last updated on 30/Jul/19
$$\int{e}^{{cos}^{−\mathrm{1}} \left({x}\right)} \:{dx} \\ $$
Answered by MJS last updated on 31/Jul/19
![∫e^(arccos x) dx= [t=arccos x → dx=−sin t dt] −∫e^t sin t dt= by parts (1) u′=e^t → u=e^t v=−sin t → v′=−cos t =−e^t sin t +∫e^t cos t dt= by parts (2) u′=e^t → u=e^t v=cos t → v′=−sin t =−e^t sin t +e^t cos t +∫e^t sin t dt now we have −∫e^t sin t dt=−e^t sin t +e^t cos t +∫e^t sin t dt ⇒ −∫e^t sin t dt=(e^t /2)(cos t −sin t) ∫e^(arccos x) dx=−e^(arccos x) (((x−(√(1−x^2 )))/2)) +C](https://www.tinkutara.com/question/Q65509.png)
$$\int\mathrm{e}^{\mathrm{arccos}\:{x}} {dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arccos}\:{x}\:\rightarrow\:{dx}=−\mathrm{sin}\:{t}\:{dt}\right] \\ $$$$−\int\mathrm{e}^{{t}} \mathrm{sin}\:{t}\:{dt}= \\ $$$$\:\:\:\:\:\mathrm{by}\:\mathrm{parts}\:\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:{u}'=\mathrm{e}^{{t}} \:\rightarrow\:{u}=\mathrm{e}^{{t}} \\ $$$$\:\:\:\:\:{v}=−\mathrm{sin}\:{t}\:\rightarrow\:{v}'=−\mathrm{cos}\:{t} \\ $$$$=−\mathrm{e}^{{t}} \mathrm{sin}\:{t}\:+\int\mathrm{e}^{{t}} \mathrm{cos}\:{t}\:{dt}= \\ $$$$\:\:\:\:\:\mathrm{by}\:\mathrm{parts}\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:{u}'=\mathrm{e}^{{t}} \:\rightarrow\:{u}=\mathrm{e}^{{t}} \\ $$$$\:\:\:\:\:{v}=\mathrm{cos}\:{t}\:\rightarrow\:{v}'=−\mathrm{sin}\:{t} \\ $$$$=−\mathrm{e}^{{t}} \mathrm{sin}\:{t}\:+\mathrm{e}^{{t}} \mathrm{cos}\:{t}\:+\int\mathrm{e}^{{t}} \mathrm{sin}\:{t}\:{dt} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have} \\ $$$$−\int\mathrm{e}^{{t}} \mathrm{sin}\:{t}\:{dt}=−\mathrm{e}^{{t}} \mathrm{sin}\:{t}\:+\mathrm{e}^{{t}} \mathrm{cos}\:{t}\:+\int\mathrm{e}^{{t}} \mathrm{sin}\:{t}\:{dt} \\ $$$$\Rightarrow\:−\int\mathrm{e}^{{t}} \mathrm{sin}\:{t}\:{dt}=\frac{\mathrm{e}^{{t}} }{\mathrm{2}}\left(\mathrm{cos}\:{t}\:−\mathrm{sin}\:{t}\right) \\ $$$$ \\ $$$$\int\mathrm{e}^{\mathrm{arccos}\:{x}} {dx}=−\mathrm{e}^{\mathrm{arccos}\:{x}} \left(\frac{{x}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{2}}\right)\:+{C} \\ $$
Answered by zakaria elghaouti last updated on 31/Jul/19
