What is the integral of 9e^{3θ}sin(3θ) ?

The integral of 9e^{3θ}sin(3θ) is 9((e^{3θ}sin(3θ))/6-(e^{3θ}cos(3θ))/6)+C

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integral of 9e^{3θ}sin(3θ)

\int 9 e^{3 θ} sin (3 θ) d θ

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

Solution steps

\int 9 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

= 9 \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

= 9 (- \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

= 9 (- \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

= 9 (- \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

9 \cdot \int e^{3 θ} sin (3 θ) d θ = 9 (- \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

= 9 (\frac{e ^{3 θ} sin ( 3 θ )}{6} - \frac{e ^{3 θ} cos ( 3 θ )}{6})

Add a constant to the solution

= 9 (\frac{e ^{3 θ} sin ( 3 θ )}{6} - \frac{e ^{3 θ} cos ( 3 θ )}{6}) + C

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What is the integral of 9e^{3θ}sin(3θ) ?
The integral of 9e^{3θ}sin(3θ) is 9((e^{3θ}sin(3θ))/6-(e^{3θ}cos(3θ))/6)+C