What is the integral of 27e^{3θ}sin(3θ) ?

The integral of 27e^{3θ}sin(3θ) is 27((e^{3θ}sin(3θ))/6-(e^{3θ}cos(3θ))/6)+C

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integral of 27e^{3θ}sin(3θ)

\int 27 e^{3 θ} sin (3 θ) d θ

27 (\frac{e ^{3 θ} sin ( 3 θ )}{6} - \frac{e ^{3 θ} cos ( 3 θ )}{6}) + C

Solution steps

\int 27 e^{3 θ} sin (3 θ) d θ

Take the constant out:

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

= 27 \cdot \int e^{3 θ} sin (3 θ) d θ

Apply Integration By Parts

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

= 27 (- \frac{1}{3} e^{3 θ} cos (3 θ) - \int - e^{3 θ} cos (3 θ) d θ)

Take the constant out:

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

= 27 (- \frac{1}{3} e^{3 θ} cos (3 θ) - (- \int e^{3 θ} cos (3 θ) d θ))

Apply Integration By Parts

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

= 27 (- \frac{1}{3} e^{3 θ} cos (3 θ) - (- (\frac{1}{3} e^{3 θ} sin (3 θ) - \int e^{3 θ} sin (3 θ) d θ)))

Therefore

27 \cdot \int e^{3 θ} sin (3 θ) d θ = 27 (- \frac{1}{3} e^{3 θ} cos (3 θ) - (- (\frac{1}{3} e^{3 θ} sin (3 θ) - \int e^{3 θ} sin (3 θ) d θ)))

Isolate

\int e^{3 θ} sin (3 θ) d θ

= 27 (\frac{e ^{3 θ} sin ( 3 θ )}{6} - \frac{e ^{3 θ} cos ( 3 θ )}{6})

Add a constant to the solution

= 27 (\frac{e ^{3 θ} sin ( 3 θ )}{6} - \frac{e ^{3 θ} cos ( 3 θ )}{6}) + C

\int 1 - csc (x) cot (x) d x

\int x 4 + 9 x^{2} d x

f (x) = \frac{x}{x ^{2} + 25}

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

f (x) = x^{2} - 3 x + 1, at x = 3

What is the integral of 27e^{3θ}sin(3θ) ?
The integral of 27e^{3θ}sin(3θ) is 27((e^{3θ}sin(3θ))/6-(e^{3θ}cos(3θ))/6)+C