# Calc.Find the derivative of the function. int(t sint dt, t=1-2x...1+2x)

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### 1 Answer

You need to use inetgration by parts formula to find the integral of function such that:

`int udv = uv - int vdu`

`u = t =gt du = dt`

`dv = sin t dt =gt v = -cos t`

`int_(1-2x)^(1+2x) t sint dt = -t cos t|_(1-2x)^(1+2x) + int_(1-2x)^(1+2x) cos t dt`

`int_(1-2x)^(1+2x) t sint dt = ((1-2x)cos(1-2x) + (1+2x)cos(1+2x)) + sin(1+2x) - sin(1-2x)`

You need to open the brackets such that:

`int_(1-2x)^(1+2x) t sint dt = cos(1-2x) - 2xcos(1-2x) + cos(1+2x) + 2xcos(1+2x) + sin(1+2x) - sin(1-2x)`

You need to differentiate the function `F(x) = cos(1-2x) - 2xcos(1-2x) + cos(1+2x) + 2xcos(1+2x) + sin(1+2x) - sin(1-2x)` with respect to x such that:

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

Reducing like terms yields:

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

`F'(x) = 2(sin(1-2x) - sin(1+2x)) - 4x(sin(1-2x) + sin(1+2x)) + 4cos(1+2x) `

**Hence, differentiating the given integral yields `(d/(dx))int_(1-2x)^(1+2x) t sint dt =2(sin(1-2x) - sin(1+2x)) - 4x(sin(1-2x) + sin(1+2x)) + 4cos(1+2x).` **