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

The integral of e^{2θ}sin(2θ) is -(e^{2θ}cos(2θ))/4+(e^{2θ}sin(2θ))/4+C

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integral of e^{2θ}sin(2θ)

\int e^{2 θ} sin (2 θ) d θ

- \frac{e ^{2 θ} cos ( 2 θ )}{4} + \frac{e ^{2 θ} sin ( 2 θ )}{4} + C

Solution steps

\int e^{2 θ} sin (2 θ) d θ

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

\int e^{2 θ} sin (2 θ) d θ

= - \frac{e ^{2 θ} cos ( 2 θ )}{4} + \frac{e ^{2 θ} sin ( 2 θ )}{4}

Add a constant to the solution

= - \frac{e ^{2 θ} cos ( 2 θ )}{4} + \frac{e ^{2 θ} sin ( 2 θ )}{4} + C

(x - 4) (x + 1)

y = \frac{6 x}{1 - tan ( x )}

\frac{\partial}{\partial x} (3 x - 2 y)

x \to 0 - lim (b x + 8)

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

What is the integral of e^{2θ}sin(2θ) ?
The integral of e^{2θ}sin(2θ) is -(e^{2θ}cos(2θ))/4+(e^{2θ}sin(2θ))/4+C