How do you use part 1 of Fundamental Theorem of Calculus to find the derivative of the function ##y=int (6+v^2)^10 dv## from sinx to cosx?

##[-(6+cos^2x)^10 sinx – (6+sin^2x)^10 cosx]##

If it is F(v) = ##int_sinx^cosx (6+v^2)##dv, it can be written as a sum of two integrals= ##int_sinx^0 (6+v^2)dv## +##int_0^cosx (6+v^2)dv##

=##int_0^cosx (6+v^2)dv-int_0^sinx (6+v^2)dv##

F'(v) =##d/dx int_0^cosx (6+v^2)dv -d/dx int_o^sinx (6+v^2)dv##

Now applying , the derivative would be ##[-(6+cos^2x)^10 sinx – (6+sin^2x)^10 cosx]##