What is the integral from 0 to x of sin(x-t)te^t ?

The integral from 0 to x of sin(x-t)te^t is (xe^x-e^x+cos(x))/2

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integral from 0 to x of sin(x-t)te^t

\int_{0}^{x} sin (x - t) t e^{t} d t

\frac{x e ^{x} - e ^{x} + cos ( x )}{2}

Solution steps

\int_{0}^{x} sin (x - t) t e^{t} d t

Apply Integration By Parts

= [t (\frac{1}{2} e^{t} cos (x - t) + \frac{1}{2} e^{t} sin (x - t)) - \int \frac{1}{2} (e^{t} cos (x - t) + e^{t} sin (x - t)) d t]_{0}^{x}

\int \frac{1}{2} (e^{t} cos (x - t) + e^{t} sin (x - t)) d t = \frac{1}{2} e^{t} cos (- t + x)

= [t (\frac{1}{2} e^{t} cos (x - t) + \frac{1}{2} e^{t} sin (x - t)) - \frac{1}{2} e^{t} cos (- t + x)]_{0}^{x}

Compute the boundaries:

x \frac{e ^{x}}{2} - \frac{e ^{x}}{2} + \frac{1}{2} cos (x)

= x \frac{e ^{x}}{2} - \frac{e ^{x}}{2} + \frac{1}{2} cos (x)

Simplify

= \frac{x e ^{x} - e ^{x} + cos ( x )}{2}

What is the integral from 0 to x of sin(x-t)te^t ?
The integral from 0 to x of sin(x-t)te^t is (xe^x-e^x+cos(x))/2