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

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

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integral of e^{3θ}sin(2θ)

\int e^{3 θ} sin (2 θ) d θ

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

Solution steps

\int e^{3 θ} sin (2 θ) d θ

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{3 θ} sin (2 θ) d θ

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

Add a constant to the solution

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

\int 1 + 4 θ^{2} d θ

e^{2 t} (t - 1)^{2}

\frac{d}{d x} ((1 - 3 x) arccoth (3 x))

y^{^{'}} + 4 x y^{3} = 0

\frac{d}{d x} (\frac{4 x ^{3}}{3} + cos (x) + 1)

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