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

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

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integral of 2e^{2θ}sin(2θ)

\int 2 e^{2 θ} sin (2 θ) d θ

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

Solution steps

\int 2 e^{2 θ} sin (2 θ) d θ

Take the constant out:

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

= 2 \cdot \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

= 2 (- \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

= 2 (- \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

= 2 (- \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

2 \cdot \int e^{2 θ} sin (2 θ) d θ = 2 (- \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

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

Add a constant to the solution

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

3 x^{\frac{3}{8}}

\frac{d}{d x} (x e^{x} + x)

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

f (x) = x^{9} ln (4 x)

\frac{d}{d x} (1. 8^{0.045 x} - 1)

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