Question

How do you use the second fundamental theorem of Calculus to find the derivative of given `int sin^3(2t-1) dt` from `[x-1, x]`?

Answer 1

Hence we need to find the derivative of

`int_(x-1)^[x] sin^3(2t-1)dt`

So we use the Leibniz Integral Rule which in our case becomes

`d/dx[int_(a(x))^(b(x)) f(t)dt]=f(b(x))*(db(x))/dx-f(a(x))*(d(a(x)))/dx`

Hence the derivative for `b(x)=x` ,`a(x)=x-1` ,`f(t)=sin^3(2t-1)`

is

`d/dx( int_(x-1)^x sin^3(2 t-1) dt) = sin^3(2x-1)*(x)'-sin^3(2(x-1)-1)*(x-1)'=sin^3(2x-1)-sin^3(2x-3)`