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`

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

`int_0^(1+2x)tsintdt=(1+2x)sin(1+2x)(2)`

`=(2+4x)(sin(1)cos(2x)+cos(1)sin(2x))`

`=2sin(1)cos(2x)+2cos(1)sin(2x)+4xsin(1)cos(2x)+4xcos(1)sin(2x)`

`-int_0^(1-2x)tsintdt=(1-2x)sin(1-2x)(-2)`

`=(2-4x)(sin(1)cos(-2x)-cos(1)sin(-2x))`

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

`=2sin(1)cos(2x)-2cos(1)sin(2x)-4xsin(1)cos(2x)+4xcos(1)sin(2x)`

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

`=(2+4x)(sin(1+2x))+(2-4x)(sin(1-2x))`

`=4sin(1)cos(2x)+8xcos(1)sin(2x)`

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