##[-(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]##
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Get The Answer 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]##
Previous answers to this question
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