What is the integral of e^{3x}cos(6x) ?

The integral of e^{3x}cos(6x) is (2e^{3x}sin(6x))/(15)+(e^{3x}cos(6x))/(15)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{3x}cos(6x)

\int e^{3 x} cos (6 x) d x

\frac{2 e ^{3 x} sin ( 6 x )}{15} + \frac{e ^{3 x} cos ( 6 x )}{15} + C

Solution steps

\int e^{3 x} cos (6 x) d x

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{3 x} cos (6 x) d x

= \frac{2 e ^{3 x} sin ( 6 x )}{15} + \frac{e ^{3 x} cos ( 6 x )}{15}

Add a constant to the solution

= \frac{2 e ^{3 x} sin ( 6 x )}{15} + \frac{e ^{3 x} cos ( 6 x )}{15} + C

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

\int sin (3 x) sin (3 x) d x

\int (5 x^{3} - 12 x^{- 3} + 9 x^{2}) d x

\int \frac{4 sin ^{3} ( x ) - 1}{sin ^{2} ( x )} d x

\int 4 x + 3^{2} + 2 d x

What is the integral of e^{3x}cos(6x) ?
The integral of e^{3x}cos(6x) is (2e^{3x}sin(6x))/(15)+(e^{3x}cos(6x))/(15)+C