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`
`=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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