find the derivative of the function. find the derivative of the function. g(x)=integral (1-2x) to (1+2x) (tsint)dt

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Find `g'(x)` if `g(x)=int_(1-2x)^(1+2x) tsint dt` :

First note that tsint is continuous, so we can apply the second part of teh fundamental theorem of calculus. However, it is only defined for integrals of the form `int_a^x` where a is a constant and x represents some function, so we have to rewrite the integral:

`int_(1-2x)^(1+2x) tsintdt=int_0^(1+2x)tsintdt+int_(1-2x)^0 tsintdt`


Now we use the second part of the fundamental theorem on each integral employing the chain rule: if `F(x)=int_a^(u(x)) f(t)dt` then `F'(x)=f(u(x))u'(x)` . So:






** Since cos(-2x)=cos(2x) and sin(-2x)=-sin(2x)**


Thus `int_0^(1+2x)tsintdt-int_0^(1-2x)tsintdt`



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