What is the integral of e^{4x}cos(7x) ?

The integral of e^{4x}cos(7x) is (7e^{4x}sin(7x))/(65)+(4e^{4x}cos(7x))/(65)+C

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integral of e^{4x}cos(7x)

\int e^{4 x} cos (7 x) d x

\frac{7 e ^{4 x} sin ( 7 x )}{65} + \frac{4 e ^{4 x} cos ( 7 x )}{65} + C

Solution steps

\int e^{4 x} cos (7 x) d x

Apply Integration By Parts

: \frac{1}{7} e^{4 x} sin (7 x) - \int \frac{4}{7} e^{4 x} sin (7 x) d x

= \frac{1}{7} e^{4 x} sin (7 x) - \int \frac{4}{7} e^{4 x} sin (7 x) d x

Take the constant out:

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

= \frac{1}{7} e^{4 x} sin (7 x) - \frac{4}{7} \cdot \int e^{4 x} sin (7 x) d x

Apply Integration By Parts

: - \frac{1}{7} e^{4 x} cos (7 x) - \int - \frac{4}{7} e^{4 x} cos (7 x) d x

= \frac{1}{7} e^{4 x} sin (7 x) - \frac{4}{7} (- \frac{1}{7} e^{4 x} cos (7 x) - \int - \frac{4}{7} e^{4 x} cos (7 x) d x)

Take the constant out:

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

= \frac{1}{7} e^{4 x} sin (7 x) - \frac{4}{7} (- \frac{1}{7} e^{4 x} cos (7 x) - (- \frac{4}{7} \cdot \int e^{4 x} cos (7 x) d x))

Therefore

\int e^{4 x} cos (7 x) d x = \frac{1}{7} e^{4 x} sin (7 x) - \frac{4}{7} (- \frac{1}{7} e^{4 x} cos (7 x) - (- \frac{4}{7} \cdot \int e^{4 x} cos (7 x) d x))

Isolate

\int e^{4 x} cos (7 x) d x

= \frac{7 e ^{4 x} sin ( 7 x )}{65} + \frac{4 e ^{4 x} cos ( 7 x )}{65}

Add a constant to the solution

= \frac{7 e ^{4 x} sin ( 7 x )}{65} + \frac{4 e ^{4 x} cos ( 7 x )}{65} + C

\frac{\partial}{\partial x} (\frac{2}{x z - z ^{4} + x ^{2}})

t^{2} \frac{d y}{d t} + y = 0, y (6) = - 3, t > 0

x \to 3 + lim (3 ln (x - 2))

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

f (x) = x^{4}, (- 2, 16)

What is the integral of e^{4x}cos(7x) ?
The integral of e^{4x}cos(7x) is (7e^{4x}sin(7x))/(65)+(4e^{4x}cos(7x))/(65)+C