What is the integral of e^{-2t}cos(3t) ?

The integral of e^{-2t}cos(3t) is (3e^{-2t}sin(3t))/(13)-(2e^{-2t}cos(3t))/(13)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{-2t}cos(3t)

\int e^{- 2 t} cos (3 t) d t

\frac{3 e ^{- 2 t} sin ( 3 t )}{13} - \frac{2 e ^{- 2 t} cos ( 3 t )}{13} + C

Solution steps

\int e^{- 2 t} cos (3 t) d t

Apply Integration By Parts

: \frac{1}{3} e^{- 2 t} sin (3 t) - \int - \frac{2}{3} e^{- 2 t} sin (3 t) d t

= \frac{1}{3} e^{- 2 t} sin (3 t) - \int - \frac{2}{3} e^{- 2 t} sin (3 t) d t

Take the constant out:

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

= \frac{1}{3} e^{- 2 t} sin (3 t) - (- \frac{2}{3} \cdot \int e^{- 2 t} sin (3 t) d t)

Apply Integration By Parts

: - \frac{1}{3} e^{- 2 t} cos (3 t) - \int \frac{2}{3} e^{- 2 t} cos (3 t) d t

= \frac{1}{3} e^{- 2 t} sin (3 t) - (- \frac{2}{3} (- \frac{1}{3} e^{- 2 t} cos (3 t) - \int \frac{2}{3} e^{- 2 t} cos (3 t) d t))

Take the constant out:

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

= \frac{1}{3} e^{- 2 t} sin (3 t) - (- \frac{2}{3} (- \frac{1}{3} e^{- 2 t} cos (3 t) - \frac{2}{3} \cdot \int e^{- 2 t} cos (3 t) d t))

Therefore

\int e^{- 2 t} cos (3 t) d t = \frac{1}{3} e^{- 2 t} sin (3 t) - (- \frac{2}{3} (- \frac{1}{3} e^{- 2 t} cos (3 t) - \frac{2}{3} \cdot \int e^{- 2 t} cos (3 t) d t))

Isolate

\int e^{- 2 t} cos (3 t) d t

= \frac{3 e ^{- 2 t} sin ( 3 t )}{13} - \frac{2 e ^{- 2 t} cos ( 3 t )}{13}

Add a constant to the solution

= \frac{3 e ^{- 2 t} sin ( 3 t )}{13} - \frac{2 e ^{- 2 t} cos ( 3 t )}{13} + C

\int (sin (θ))^{4} d θ

\int \frac{cos ( 2 x )}{1 - cos ^{2} ( 2 x )} d x

\int 8 - x^{3} d x

\int \frac{x ^{2} + 1}{x ^{2} + 2 x + 2} d x

\int 5 x^{3} + 3 x - 2 (5 x^{2} + 1) d x

What is the integral of e^{-2t}cos(3t) ?
The integral of e^{-2t}cos(3t) is (3e^{-2t}sin(3t))/(13)-(2e^{-2t}cos(3t))/(13)+C