Question #e63af

Question

↳Redirected from "How do you evaluate #sin^-1 (-sqrt3/2)#?"

2351 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-04T11:19:29

The answer #==-73/504=-0.14484#

We need

#intx^ndx=x^(n+1)/(n+1)+C(x!=-1)#

Therefore,

#int_0^1x^6(4x^2+x-5)dx#

#=int_0^1(4x^8+x^7-5x^6)dx#

#=[4x^9/9+x^8/8-5x^7/7] _0^1#

#=(4/9+1/8-5/7)-(0)#

#=(224+63-360)/504#

#=-73/504=-0.14484#