What is the integral of e^{5θ}sin(6θ) ?

The integral of e^{5θ}sin(6θ) is -(6e^{5θ}cos(6θ))/(61)+(5e^{5θ}sin(6θ))/(61)+C

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integral of e^{5θ}sin(6θ)

\int e^{5 θ} sin (6 θ) d θ

- \frac{6 e ^{5 θ} cos ( 6 θ )}{61} + \frac{5 e ^{5 θ} sin ( 6 θ )}{61} + C

Solution steps

\int e^{5 θ} sin (6 θ) d θ

Apply Integration By Parts

= - \frac{1}{6} e^{5 θ} cos (6 θ) - \int - \frac{5}{6} e^{5 θ} cos (6 θ) d θ

Take the constant out:

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

= - \frac{1}{6} e^{5 θ} cos (6 θ) - (- \frac{5}{6} \cdot \int e^{5 θ} cos (6 θ) d θ)

Apply Integration By Parts

= - \frac{1}{6} e^{5 θ} cos (6 θ) - (- \frac{5}{6} (\frac{1}{6} e^{5 θ} sin (6 θ) - \int \frac{5}{6} e^{5 θ} sin (6 θ) d θ))

Take the constant out:

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

= - \frac{1}{6} e^{5 θ} cos (6 θ) - (- \frac{5}{6} (\frac{1}{6} e^{5 θ} sin (6 θ) - \frac{5}{6} \cdot \int e^{5 θ} sin (6 θ) d θ))

Therefore

\int e^{5 θ} sin (6 θ) d θ = - \frac{1}{6} e^{5 θ} cos (6 θ) - (- \frac{5}{6} (\frac{1}{6} e^{5 θ} sin (6 θ) - \frac{5}{6} \cdot \int e^{5 θ} sin (6 θ) d θ))

Isolate

\int e^{5 θ} sin (6 θ) d θ

= - \frac{6 e ^{5 θ} cos ( 6 θ )}{61} + \frac{5 e ^{5 θ} sin ( 6 θ )}{61}

Add a constant to the solution

= - \frac{6 e ^{5 θ} cos ( 6 θ )}{61} + \frac{5 e ^{5 θ} sin ( 6 θ )}{61} + C

What is the integral of e^{5θ}sin(6θ) ?
The integral of e^{5θ}sin(6θ) is -(6e^{5θ}cos(6θ))/(61)+(5e^{5θ}sin(6θ))/(61)+C