Question

Help please!

Use the Fundamental Theorem of Calculus to find the derivative of...

h(x)= integral of -3 to sin(x) (cos(t^3)+sin(t))

Answer 1

I know i have to use the rule h(x)= integral of a to u(x) g(t) dt equals h'(x)= u'(x)g(u(x))

Answer 2

let f(x)=cos(t^3)+sin(t)
then h(x)=F(sin(x))-F(-3)
then h'(x)=d/dx(F(sin(x)))-0=F'(sin(x))*sin'(x)=f(sin(x))*cos(x)

Answer 3

(F is the antiderivative of f)

Answer 4

right so that would leave the equation to be...

(cos(sin^3(x))+sin(x))+(cos(-3)+(-3))

Answer 5

f(sin(x))=cos(sin^3(x))+sin(sin(x))
f(sin(x))*cos(x)=cos(sin^3(x))*cos(x)+sin(sin(x))*cos(x)

Answer 6

i am so confused :(

Answer 7

using the rule h(x) = integral of a to u(x) g(t) dt equals h'(x)= u'(x)g(u(x))
a=-3
u(x)=sin(x)
g(t)=cos(t^3)+sin(t)
now find h(x)

Answer 8

*now find h'(x)

Answer 9

h'(x)= u'(x)g(u(x))
u'(x)=cos(x)
g(u(x))=g(sin(x))=cos(sin(x)^3)+sin(sin(x))
h'(x)=u'(x)g(u(x))=(cos(sin(x)^3)+sin(sin(x))) * cos(x)

Answer 10

where did the extra sin for g(u(x)) come from?

Answer 11

I just have no idea what to do :(