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

1 Answer
Nov 7, 2016

Answer:

# 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)))#