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

The integral of+e^{-3x}sin(bx) is -(be^{-3x}cos(bx))/(b^2+9)-(3e^{-3x}sin(bx))/(b^2+9)+C

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

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

- \frac{b e ^{- 3 x} cos ( b x )}{b ^{2} + 9} - \frac{3 e ^{- 3 x} sin ( b x )}{b ^{2} + 9} + C

Solution steps

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

Apply Integration By Parts

= - e^{- 3 x} \frac{1}{b} cos (b x) - \int e^{- 3 x} \frac{3}{b} cos (b x) d x

Take the constant out:

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

= - e^{- 3 x} \frac{1}{b} cos (b x) - \frac{3}{b} \cdot \int e^{- 3 x} cos (b x) d x

Apply Integration By Parts

= - e^{- 3 x} \frac{1}{b} cos (b x) - \frac{3}{b} (e^{- 3 x} \frac{1}{b} sin (b x) - \int - e^{- 3 x} \frac{3}{b} sin (b x) d x)

Take the constant out:

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

= - e^{- 3 x} \frac{1}{b} cos (b x) - \frac{3}{b} (e^{- 3 x} \frac{1}{b} sin (b x) - (- \frac{3}{b} \cdot \int e^{- 3 x} sin (b x) d x))

Therefore

\int e^{- 3 x} sin (b x) d x = - e^{- 3 x} \frac{1}{b} cos (b x) - \frac{3}{b} (e^{- 3 x} \frac{1}{b} sin (b x) - (- \frac{3}{b} \cdot \int e^{- 3 x} sin (b x) d x))

Isolate

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

= - \frac{b e ^{- 3 x} cos ( b x )}{b ^{2} + 9} - \frac{3 e ^{- 3 x} sin ( b x )}{b ^{2} + 9}

Add a constant to the solution

= - \frac{b e ^{- 3 x} cos ( b x )}{b ^{2} + 9} - \frac{3 e ^{- 3 x} sin ( b x )}{b ^{2} + 9} + C

\frac{d}{d x} (arcsin (\frac{x}{3}))

\int \frac{4 x + 1}{4 x} d x

n = 1 \sum \infty \frac{1}{n ^{2}} - \frac{1}{n}

x \to 5 lim (x^{2} - 25)

\int \frac{2}{( x ^{2} - 3 x + 2 )} d x

What is the integral of+e^{-3x}sin(bx) ?
The integral of+e^{-3x}sin(bx) is -(be^{-3x}cos(bx))/(b^2+9)-(3e^{-3x}sin(bx))/(b^2+9)+C