What is the integral from 0 to t of e^{-2t}sin(5t) ?

The integral from 0 to t of e^{-2t}sin(5t) is (-5e^{-2t}cos(5t)-2e^{-2t}sin(5t)+5)/(29)

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integral from 0 to t of e^{-2t}sin(5t)

\int_{0}^{t} e^{- 2 t} sin (5 t) d t

\frac{- 5 e ^{- 2 t} cos ( 5 t ) - 2 e ^{- 2 t} sin ( 5 t ) + 5}{29}

Solution steps

\int_{0}^{t} e^{- 2 t} sin (5 t) d t

Compute the indefinite integral:

\int e^{- 2 t} sin (5 t) d t = - \frac{5 e ^{- 2 t} cos ( 5 t )}{29} - \frac{2 e ^{- 2 t} sin ( 5 t )}{29} + C

Compute the boundaries:

\int_{0}^{t} e^{- 2 t} sin (5 t) d t = - \frac{5 e ^{- 2 t} cos ( 5 t )}{29} - \frac{2 e ^{- 2 t} sin ( 5 t )}{29} - (- \frac{5}{29})

= - \frac{5 e ^{- 2 t} cos ( 5 t )}{29} - \frac{2 e ^{- 2 t} sin ( 5 t )}{29} - (- \frac{5}{29})

Simplify

= \frac{- 5 e ^{- 2 t} cos ( 5 t ) - 2 e ^{- 2 t} sin ( 5 t ) + 5}{29}

\frac{d}{d x} (9 cos^{3} (x))

\int_{0}^{\frac{1}{6}} x ln (6 x) d x

y = x + 1, (- 1, 4)

x \to \infty lim (i^{x})

f (x) = (x^{3} + 5) (x^{3} + x + 1)

What is the integral from 0 to t of e^{-2t}sin(5t) ?
The integral from 0 to t of e^{-2t}sin(5t) is (-5e^{-2t}cos(5t)-2e^{-2t}sin(5t)+5)/(29)