What is the integral of 10e^{-30t}sin(40t) ?

The integral of 10e^{-30t}sin(40t) is 10(-(2e^{-30t}cos(40t))/(125)-(3e^{-30t}sin(40t))/(250))+C

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integral of 10e^{-30t}sin(40t)

\int 10 e^{- 30 t} sin (40 t) d t

10 (- \frac{2 e ^{- 30 t} cos ( 40 t )}{125} - \frac{3 e ^{- 30 t} sin ( 40 t )}{250}) + C

Solution steps

\int 10 e^{- 30 t} sin (40 t) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \int e^{- 30 t} sin (40 t) d t

Apply Integration By Parts

: - \frac{1}{40} e^{- 30 t} cos (40 t) - \int \frac{3}{4} e^{- 30 t} cos (40 t) d t

= 10 (- \frac{1}{40} e^{- 30 t} cos (40 t) - \int \frac{3}{4} e^{- 30 t} cos (40 t) d t)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 (- \frac{1}{40} e^{- 30 t} cos (40 t) - \frac{3}{4} \cdot \int e^{- 30 t} cos (40 t) d t)

Apply Integration By Parts

: \frac{1}{40} e^{- 30 t} sin (40 t) - \int - \frac{3}{4} e^{- 30 t} sin (40 t) d t

= 10 (- \frac{1}{40} e^{- 30 t} cos (40 t) - \frac{3}{4} (\frac{1}{40} e^{- 30 t} sin (40 t) - \int - \frac{3}{4} e^{- 30 t} sin (40 t) d t))

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 (- \frac{1}{40} e^{- 30 t} cos (40 t) - \frac{3}{4} (\frac{1}{40} e^{- 30 t} sin (40 t) - (- \frac{3}{4} \cdot \int e^{- 30 t} sin (40 t) d t)))

Therefore

10 \cdot \int e^{- 30 t} sin (40 t) d t = 10 (- \frac{1}{40} e^{- 30 t} cos (40 t) - \frac{3}{4} (\frac{1}{40} e^{- 30 t} sin (40 t) - (- \frac{3}{4} \cdot \int e^{- 30 t} sin (40 t) d t)))

Isolate

\int e^{- 30 t} sin (40 t) d t

= 10 (- \frac{2 e ^{- 30 t} cos ( 40 t )}{125} - \frac{3 e ^{- 30 t} sin ( 40 t )}{250})

Add a constant to the solution

= 10 (- \frac{2 e ^{- 30 t} cos ( 40 t )}{125} - \frac{3 e ^{- 30 t} sin ( 40 t )}{250}) + C

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What is the integral of 10e^{-30t}sin(40t) ?
The integral of 10e^{-30t}sin(40t) is 10(-(2e^{-30t}cos(40t))/(125)-(3e^{-30t}sin(40t))/(250))+C