# How do you use the second fundamental theorem of Calculus to find the derivative of given int (1)/(2+sin(t)) from [0, lnx]?

## d/dxint_0^(lnx) 1/(2+sint)dt = 1/(x(2+sin(lnx)))##

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How do you use the second fundamental theorem of Calculus to find the derivative of given int (1)/(2+sin(t)) from [0, lnx]?
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The second fundamental theorem of calculus states that if F is any anti-derivative of f, then

## int_a^b f(t)dt = F(b) – F(a) ##, and ## F'(x)=f(x) ##

Note that as this is a definite integral, the RHS is a number.

If ## f(x) = 1/(2+sint) ## and ## F'(x)=f(x) ## Then

## int_0^(lnx) 1/(2+sint)dt=F(lnx) – F(0) ##

If we differentiate wrt ##x## we get

## d/dxint_0^(lnx) 1/(2+sint)dt = d/dx(F(lnx) – F(0)) ##
## :. d/dxint_0^(lnx) 1/(2+sint)dt = d/dxF(lnx) ## (as ##F(0)## is a constant)
## :. d/dxint_0^(lnx) 1/(2+sint)dt = d/dx(lnx)F'(lnx) ## (using the chain rule)
## :. d/dxint_0^(lnx) 1/(2+sint)dt = 1/x* 1/(2+sin(lnx))##
## :. d/dxint_0^(lnx) 1/(2+sint)dt = 1/(x(2+sin(lnx)))##

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