Question

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

Answer 1

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

Explanation:

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)))`