Solve ∫ (from 0 to 1) of e^-xiomega(t-tau)sin(omega_{d}(t-tau))dtau

$$\int _ { 0 } ^ { 1 } e ^ { - \xi \omega ( t - \tau ) } \sin ( \omega _ { d } ( t - \tau ) ) d \tau$$

Evaluate

$$\frac{\sin(\omega _{d}\left(t-2\pi \right))}{e^{\xi \omega \left(t-2\pi \right)}}$$

Solution Steps

Substitute 2 * \pi for \tau.

$$\int_{0}^{1} {e ^ {-\xi \omega {(t - 2 * \pi)}} \sin(\omega_{d} {(t - 2 * \pi)})} d\tau$$

Evaluate the indefinite integral first.

$$\int \frac{\sin(\omega _{d}\left(t-2\pi \right))}{e^{\xi \omega \left(t-2\pi \right)}}\mathrm{d}\tau$$

Find the integral of \frac{\sin(\omega _{d}\left(t-2\pi \right))}{e^{\xi \omega \left(t-2\pi \right)}} using the table of common integrals rule \int a\mathrm{d}\tau =a\tau .

$$\frac{\sin(\omega _{d}\left(t-2\pi \right))}{e^{\xi \omega \left(t-2\pi \right)}}\tau$$

Simplify.

$$\frac{\sin(\omega _{d}\left(t-2\pi \right))\tau }{e^{\xi \omega \left(t-2\pi \right)}}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$e^{-\xi \omega \left(t-2\pi \right)}\sin(\omega _{d}\left(t-2\pi \right))+0e^{-\xi \omega \left(t-2\pi \right)}\sin(\omega _{d}\left(t-2\pi \right))$$

Simplify.

$$\frac{\sin(\omega _{d}\left(t-2\pi \right))}{e^{\xi \omega \left(t-2\pi \right)}}$$

Differentiate w.r.t. ξ

$$-\frac{\omega \left(t-2\pi \right)\sin(\omega _{d}\left(t-2\pi \right))}{e^{\xi \omega \left(t-2\pi \right)}}$$