What is the integral of e^{-2θ}sin(3θ) ?

The integral of e^{-2θ}sin(3θ) is -(3e^{-2θ}cos(3θ))/(13)-(2e^{-2θ}sin(3θ))/(13)+C

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integral of e^{-2θ}sin(3θ)

\int e^{- 2 θ} sin (3 θ) d θ

- \frac{3 e ^{- 2 θ} cos ( 3 θ )}{13} - \frac{2 e ^{- 2 θ} sin ( 3 θ )}{13} + C

Solution steps

\int e^{- 2 θ} sin (3 θ) d θ

Apply Integration By Parts

: - \frac{1}{3} e^{- 2 θ} cos (3 θ) - \int \frac{2}{3} e^{- 2 θ} cos (3 θ) d θ

= - \frac{1}{3} e^{- 2 θ} cos (3 θ) - \int \frac{2}{3} e^{- 2 θ} cos (3 θ) d θ

Take the constant out:

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

= - \frac{1}{3} e^{- 2 θ} cos (3 θ) - \frac{2}{3} \cdot \int e^{- 2 θ} cos (3 θ) d θ

Apply Integration By Parts

: \frac{1}{3} e^{- 2 θ} sin (3 θ) - \int - \frac{2}{3} e^{- 2 θ} sin (3 θ) d θ

= - \frac{1}{3} e^{- 2 θ} cos (3 θ) - \frac{2}{3} (\frac{1}{3} e^{- 2 θ} sin (3 θ) - \int - \frac{2}{3} e^{- 2 θ} sin (3 θ) d θ)

Take the constant out:

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

= - \frac{1}{3} e^{- 2 θ} cos (3 θ) - \frac{2}{3} (\frac{1}{3} e^{- 2 θ} sin (3 θ) - (- \frac{2}{3} \cdot \int e^{- 2 θ} sin (3 θ) d θ))

Therefore

\int e^{- 2 θ} sin (3 θ) d θ = - \frac{1}{3} e^{- 2 θ} cos (3 θ) - \frac{2}{3} (\frac{1}{3} e^{- 2 θ} sin (3 θ) - (- \frac{2}{3} \cdot \int e^{- 2 θ} sin (3 θ) d θ))

Isolate

\int e^{- 2 θ} sin (3 θ) d θ

= - \frac{3 e ^{- 2 θ} cos ( 3 θ )}{13} - \frac{2 e ^{- 2 θ} sin ( 3 θ )}{13}

Add a constant to the solution

= - \frac{3 e ^{- 2 θ} cos ( 3 θ )}{13} - \frac{2 e ^{- 2 θ} sin ( 3 θ )}{13} + C

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What is the integral of e^{-2θ}sin(3θ) ?
The integral of e^{-2θ}sin(3θ) is -(3e^{-2θ}cos(3θ))/(13)-(2e^{-2θ}sin(3θ))/(13)+C