Question Number 221957 by fantastic last updated on 13/Jun/25
$$\int\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\:{x}\right){dx} \\ $$
Answered by Ghisom last updated on 13/Jun/25
![∫arcsin cos x dx= [by parts: { ((u′=1 → u=x)),((v=arcsin cos x → v′=−σ)) :}] where σ=sign (sin x) =xarcsin cos x +σ ∫xdx= =((σx^2 )/2)+xarcsin cos x +C](https://www.tinkutara.com/question/Q221961.png)
$$\int\mathrm{arcsin}\:\mathrm{cos}\:{x}\:{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}:\:\begin{cases}{{u}'=\mathrm{1}\:\rightarrow\:{u}={x}}\\{{v}=\mathrm{arcsin}\:\mathrm{cos}\:{x}\:\rightarrow\:{v}'=−\sigma}\end{cases}\right] \\ $$$$\:\:\:\:\:\mathrm{where}\:\sigma=\mathrm{sign}\:\left(\mathrm{sin}\:{x}\right) \\ $$$$={x}\mathrm{arcsin}\:\mathrm{cos}\:{x}\:+\sigma\:\int{xdx}= \\ $$$$=\frac{\sigma{x}^{\mathrm{2}} }{\mathrm{2}}+{x}\mathrm{arcsin}\:\mathrm{cos}\:{x}\:+{C} \\ $$
Answered by MathematicalUser2357 last updated on 15/Jun/25
$${Blue}:\:{Original},\:{Red}:\:{Final} \\ $$$$ \\ $$
Commented by MathematicalUser2357 last updated on 15/Jun/25
Sometimes the graph color won't work