What is the integral of cos(2t)(e^{6t}) ?

The integral of cos(2t)(e^{6t}) is (e^{6t}sin(2t))/(20)+(3e^{6t}cos(2t))/(20)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of cos(2t)(e^{6t})

\int cos (2 t) (e^{6 t}) d t

\frac{e ^{6 t} sin ( 2 t )}{20} + \frac{3 e ^{6 t} cos ( 2 t )}{20} + C

Solution steps

\int cos (2 t) e^{6 t} d t

Apply Integration By Parts

: \frac{1}{2} e^{6 t} sin (2 t) - \int 3 e^{6 t} sin (2 t) d t

= \frac{1}{2} e^{6 t} sin (2 t) - \int 3 e^{6 t} sin (2 t) d t

Take the constant out:

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

= \frac{1}{2} e^{6 t} sin (2 t) - 3 \cdot \int e^{6 t} sin (2 t) d t

Apply Integration By Parts

: - \frac{1}{2} e^{6 t} cos (2 t) - \int - 3 e^{6 t} cos (2 t) d t

= \frac{1}{2} e^{6 t} sin (2 t) - 3 (- \frac{1}{2} e^{6 t} cos (2 t) - \int - 3 e^{6 t} cos (2 t) d t)

Take the constant out:

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

= \frac{1}{2} e^{6 t} sin (2 t) - 3 (- \frac{1}{2} e^{6 t} cos (2 t) - (- 3 \cdot \int e^{6 t} cos (2 t) d t))

Therefore

\int e^{6 t} cos (2 t) d t = \frac{1}{2} e^{6 t} sin (2 t) - 3 (- \frac{1}{2} e^{6 t} cos (2 t) - (- 3 \cdot \int e^{6 t} cos (2 t) d t))

Isolate

\int e^{6 t} cos (2 t) d t

= \frac{e ^{6 t} sin ( 2 t )}{20} + \frac{3 e ^{6 t} cos ( 2 t )}{20}

Add a constant to the solution

= \frac{e ^{6 t} sin ( 2 t )}{20} + \frac{3 e ^{6 t} cos ( 2 t )}{20} + C

\int \frac{5 x ^{4}}{x ^{5} + 2} d x

\int \frac{cos ( t )}{2 sin ( t ) - 1} d t

\int x^{3} + 2 x^{2} + x d x

\int (e^{x} - e^{- x} + e^{3 x}) d x

\int \frac{3}{( 2 x ^{2} + 10 x + 4 ) ^{\frac{7}{2}}} d x

What is the integral of cos(2t)(e^{6t}) ?
The integral of cos(2t)(e^{6t}) is (e^{6t}sin(2t))/(20)+(3e^{6t}cos(2t))/(20)+C