What is #int ((arcsinx)^9) / (sqrt(1-x^2) dx#?

Question

↳Redirected from "How does a vector quantity differ from a scalar quantity?"

3385 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-29T02:07:16

#color(blue)[int ((arcsinx)^9) / (sqrt(1-x^2)] dx=(arcsinx)^10/10+c]#

#int ((arcsinx)^9) / (sqrt(1-x^2)] dx#

lets suppose:

#u=arcsinx#

#du=1/sqrt(1-x^2)*dx#

#dx=sqrt(1-x^2)*du#

#int ((arcsinx)^9) / (sqrt(1-x^2) dx]=int ((u)^9*sqrt(1-x^2))/ (sqrt(1-x^2)]*du#

#intu^9*du=u^10/10=(arcsinx)^10/10+c#