What is the integral of 5e^tsin(2t) ?

The integral of 5e^tsin(2t) is -2e^tcos(2t)+e^tsin(2t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 5e^tsin(2t)

\int 5 e^{t} sin (2 t) d t

- 2 e^{t} cos (2 t) + e^{t} sin (2 t) + C

Solution steps

\int 5 e^{t} sin (2 t) d t

Take the constant out:

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

= 5 \cdot \int e^{t} sin (2 t) d t

Apply Integration By Parts

= 5 (- \frac{1}{2} e^{t} cos (2 t) - \int - \frac{1}{2} e^{t} cos (2 t) d t)

Take the constant out:

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

= 5 (- \frac{1}{2} e^{t} cos (2 t) - (- \frac{1}{2} \cdot \int e^{t} cos (2 t) d t))

Apply Integration By Parts

= 5 (- \frac{1}{2} e^{t} cos (2 t) - (- \frac{1}{2} (e^{t} \frac{1}{2} sin (2 t) - \int e^{t} \frac{1}{2} sin (2 t) d t)))

Take the constant out:

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

= 5 (- \frac{1}{2} e^{t} cos (2 t) - (- \frac{1}{2} (e^{t} \frac{1}{2} sin (2 t) - \frac{1}{2} \cdot \int e^{t} sin (2 t) d t)))

Therefore

5 \cdot \int e^{t} sin (2 t) d t = 5 (- \frac{1}{2} e^{t} cos (2 t) - (- \frac{1}{2} (e^{t} \frac{1}{2} sin (2 t) - \frac{1}{2} \cdot \int e^{t} sin (2 t) d t)))

Isolate

\int e^{t} sin (2 t) d t

= 5 (- \frac{2 e ^{t} cos ( 2 t )}{5} + \frac{e ^{t} sin ( 2 t )}{5})

Simplify

5 (- \frac{2 e ^{t} cos ( 2 t )}{5} + \frac{e ^{t} sin ( 2 t )}{5}) : - 2 e^{t} cos (2 t) + e^{t} sin (2 t)

= - 2 e^{t} cos (2 t) + e^{t} sin (2 t)

Add a constant to the solution

= - 2 e^{t} cos (2 t) + e^{t} sin (2 t) + C

What is the integral of 5e^tsin(2t) ?
The integral of 5e^tsin(2t) is -2e^tcos(2t)+e^tsin(2t)+C