What is the integral of e^{8θ}sin(9θ) ?

The integral of e^{8θ}sin(9θ) is -(9e^{8θ}cos(9θ))/(145)+(8e^{8θ}sin(9θ))/(145)+C

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integral of e^{8θ}sin(9θ)

\int e^{8 θ} sin (9 θ) d θ

- \frac{9 e ^{8 θ} cos ( 9 θ )}{145} + \frac{8 e ^{8 θ} sin ( 9 θ )}{145} + C

Solution steps

\int e^{8 θ} sin (9 θ) d θ

Apply Integration By Parts

= - \frac{1}{9} e^{8 θ} cos (9 θ) - \int - \frac{8}{9} e^{8 θ} cos (9 θ) d θ

Take the constant out:

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

= - \frac{1}{9} e^{8 θ} cos (9 θ) - (- \frac{8}{9} \cdot \int e^{8 θ} cos (9 θ) d θ)

Apply Integration By Parts

= - \frac{1}{9} e^{8 θ} cos (9 θ) - (- \frac{8}{9} (\frac{1}{9} e^{8 θ} sin (9 θ) - \int \frac{8}{9} e^{8 θ} sin (9 θ) d θ))

Take the constant out:

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

= - \frac{1}{9} e^{8 θ} cos (9 θ) - (- \frac{8}{9} (\frac{1}{9} e^{8 θ} sin (9 θ) - \frac{8}{9} \cdot \int e^{8 θ} sin (9 θ) d θ))

Therefore

\int e^{8 θ} sin (9 θ) d θ = - \frac{1}{9} e^{8 θ} cos (9 θ) - (- \frac{8}{9} (\frac{1}{9} e^{8 θ} sin (9 θ) - \frac{8}{9} \cdot \int e^{8 θ} sin (9 θ) d θ))

Isolate

\int e^{8 θ} sin (9 θ) d θ

= - \frac{9 e ^{8 θ} cos ( 9 θ )}{145} + \frac{8 e ^{8 θ} sin ( 9 θ )}{145}

Add a constant to the solution

= - \frac{9 e ^{8 θ} cos ( 9 θ )}{145} + \frac{8 e ^{8 θ} sin ( 9 θ )}{145} + C

What is the integral of e^{8θ}sin(9θ) ?
The integral of e^{8θ}sin(9θ) is -(9e^{8θ}cos(9θ))/(145)+(8e^{8θ}sin(9θ))/(145)+C