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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

Take the constant out:

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

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

Apply Integration By Parts

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

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