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

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Answer 1 · 2016-03-05T21:33:17

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

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

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