What is the integral of 3e^{-x}sin(x) ?

The integral of 3e^{-x}sin(x) is 3(-(e^{-x}cos(x))/2-(e^{-x}sin(x))/2)+C

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integral of 3e^{-x}sin(x)

\int 3 e^{- x} sin (x) d x

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

Solution steps

\int 3 e^{- x} sin (x) d x

Take the constant out:

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

= 3 \cdot \int e^{- x} sin (x) d x

Apply Integration By Parts

: - e^{- x} cos (x) - \int e^{- x} cos (x) d x

= 3 (- e^{- x} cos (x) - \int e^{- x} cos (x) d x)

Apply Integration By Parts

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

= 3 (- e^{- x} cos (x) - (e^{- x} sin (x) - \int - e^{- x} sin (x) d x))

Take the constant out:

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

= 3 (- e^{- x} cos (x) - (e^{- x} sin (x) - (- \int e^{- x} sin (x) d x)))

Therefore

3 \cdot \int e^{- x} sin (x) d x = 3 (- e^{- x} cos (x) - (e^{- x} sin (x) - (- \int e^{- x} sin (x) d x)))

Isolate

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

= 3 (- \frac{e ^{- x} cos ( x )}{2} - \frac{e ^{- x} sin ( x )}{2})

Add a constant to the solution

= 3 (- \frac{e ^{- x} cos ( x )}{2} - \frac{e ^{- x} sin ( x )}{2}) + C

\frac{2 s - 4}{s ^{2} - 4 s + 13}

\frac{d}{d x} (- 0.5 x^{2} + 64 x - 25)

y^{^{''}} + y = 3 sin (x)

y^{^{''}} + y^{^{'}} = 6 sin (2 x)

x \to 8 lim (\frac{5 x - 2}{x + 8})

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