What is the integral of e^{3θ}sin(4θ) ?

The integral of e^{3θ}sin(4θ) is -(4e^{3θ}cos(4θ))/(25)+(3e^{3θ}sin(4θ))/(25)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{3θ}sin(4θ)

\int e^{3 θ} sin (4 θ) d θ

- \frac{4 e ^{3 θ} cos ( 4 θ )}{25} + \frac{3 e ^{3 θ} sin ( 4 θ )}{25} + C

Solution steps

\int e^{3 θ} sin (4 θ) d θ

Apply Integration By Parts

: - \frac{1}{4} e^{3 θ} cos (4 θ) - \int - \frac{3}{4} e^{3 θ} cos (4 θ) d θ

= - \frac{1}{4} e^{3 θ} cos (4 θ) - \int - \frac{3}{4} e^{3 θ} cos (4 θ) d θ

Take the constant out:

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

= - \frac{1}{4} e^{3 θ} cos (4 θ) - (- \frac{3}{4} \cdot \int e^{3 θ} cos (4 θ) d θ)

Apply Integration By Parts

: \frac{1}{4} e^{3 θ} sin (4 θ) - \int \frac{3}{4} e^{3 θ} sin (4 θ) d θ

= - \frac{1}{4} e^{3 θ} cos (4 θ) - (- \frac{3}{4} (\frac{1}{4} e^{3 θ} sin (4 θ) - \int \frac{3}{4} e^{3 θ} sin (4 θ) d θ))

Take the constant out:

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

= - \frac{1}{4} e^{3 θ} cos (4 θ) - (- \frac{3}{4} (\frac{1}{4} e^{3 θ} sin (4 θ) - \frac{3}{4} \cdot \int e^{3 θ} sin (4 θ) d θ))

Therefore

\int e^{3 θ} sin (4 θ) d θ = - \frac{1}{4} e^{3 θ} cos (4 θ) - (- \frac{3}{4} (\frac{1}{4} e^{3 θ} sin (4 θ) - \frac{3}{4} \cdot \int e^{3 θ} sin (4 θ) d θ))

Isolate

\int e^{3 θ} sin (4 θ) d θ

= - \frac{4 e ^{3 θ} cos ( 4 θ )}{25} + \frac{3 e ^{3 θ} sin ( 4 θ )}{25}

Add a constant to the solution

= - \frac{4 e ^{3 θ} cos ( 4 θ )}{25} + \frac{3 e ^{3 θ} sin ( 4 θ )}{25} + C

\int_{0}^{2} (x - 7)^{2} d x

\int 3 a x d x

x \to 3 - lim (\frac{1}{x ^{2} - 9})

x \to 2 lim (\frac{4 - x ^{2}}{x})

y = x + x, (16, 20)

What is the integral of e^{3θ}sin(4θ) ?
The integral of e^{3θ}sin(4θ) is -(4e^{3θ}cos(4θ))/(25)+(3e^{3θ}sin(4θ))/(25)+C