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)))##

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)))##