What is the integral of e^{4θ}sin(5θ) ?

The integral of e^{4θ}sin(5θ) is -(5e^{4θ}cos(5θ))/(41)+(4e^{4θ}sin(5θ))/(41)+C

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integral of e^{4θ}sin(5θ)

\int e^{4 θ} sin (5 θ) d θ

- \frac{5 e ^{4 θ} cos ( 5 θ )}{41} + \frac{4 e ^{4 θ} sin ( 5 θ )}{41} + C

Solution steps

\int e^{4 θ} sin (5 θ) d θ

Apply Integration By Parts

: - \frac{1}{5} e^{4 θ} cos (5 θ) - \int - \frac{4}{5} e^{4 θ} cos (5 θ) d θ

= - \frac{1}{5} e^{4 θ} cos (5 θ) - \int - \frac{4}{5} e^{4 θ} cos (5 θ) d θ

Take the constant out:

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

= - \frac{1}{5} e^{4 θ} cos (5 θ) - (- \frac{4}{5} \cdot \int e^{4 θ} cos (5 θ) d θ)

Apply Integration By Parts

: \frac{1}{5} e^{4 θ} sin (5 θ) - \int \frac{4}{5} e^{4 θ} sin (5 θ) d θ

= - \frac{1}{5} e^{4 θ} cos (5 θ) - (- \frac{4}{5} (\frac{1}{5} e^{4 θ} sin (5 θ) - \int \frac{4}{5} e^{4 θ} sin (5 θ) d θ))

Take the constant out:

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

= - \frac{1}{5} e^{4 θ} cos (5 θ) - (- \frac{4}{5} (\frac{1}{5} e^{4 θ} sin (5 θ) - \frac{4}{5} \cdot \int e^{4 θ} sin (5 θ) d θ))

Therefore

\int e^{4 θ} sin (5 θ) d θ = - \frac{1}{5} e^{4 θ} cos (5 θ) - (- \frac{4}{5} (\frac{1}{5} e^{4 θ} sin (5 θ) - \frac{4}{5} \cdot \int e^{4 θ} sin (5 θ) d θ))

Isolate

\int e^{4 θ} sin (5 θ) d θ

= - \frac{5 e ^{4 θ} cos ( 5 θ )}{41} + \frac{4 e ^{4 θ} sin ( 5 θ )}{41}

Add a constant to the solution

= - \frac{5 e ^{4 θ} cos ( 5 θ )}{41} + \frac{4 e ^{4 θ} sin ( 5 θ )}{41} + C

f (x) = 3 x^{2}, g (x) = - 3 x + 6, [- 4, 2]

4 e^{x} - 3 x^{2}

x \to 48 lim (\frac{x ^{2}}{8} - \frac{6}{x})

\int x^{2} \cdot e^{4 x} d x

f (x) = \frac{1}{x + 2}, at x = 9

What is the integral of e^{4θ}sin(5θ) ?
The integral of e^{4θ}sin(5θ) is -(5e^{4θ}cos(5θ))/(41)+(4e^{4θ}sin(5θ))/(41)+C