What is the integral of e^{4x}cos(3x) ?

The integral of e^{4x}cos(3x) is (3e^{4x}sin(3x))/(25)+(4e^{4x}cos(3x))/(25)+C

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integral of e^{4x}cos(3x)

\int e^{4 x} cos (3 x) d x

\frac{3 e ^{4 x} sin ( 3 x )}{25} + \frac{4 e ^{4 x} cos ( 3 x )}{25} + C

Solution steps

\int e^{4 x} cos (3 x) d x

Apply Integration By Parts

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

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

Take the constant out:

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

= \frac{1}{3} e^{4 x} sin (3 x) - \frac{4}{3} \cdot \int e^{4 x} sin (3 x) d x

Apply Integration By Parts

: - \frac{1}{3} e^{4 x} cos (3 x) - \int - \frac{4}{3} e^{4 x} cos (3 x) d x

= \frac{1}{3} e^{4 x} sin (3 x) - \frac{4}{3} (- \frac{1}{3} e^{4 x} cos (3 x) - \int - \frac{4}{3} e^{4 x} cos (3 x) d x)

Take the constant out:

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

= \frac{1}{3} e^{4 x} sin (3 x) - \frac{4}{3} (- \frac{1}{3} e^{4 x} cos (3 x) - (- \frac{4}{3} \cdot \int e^{4 x} cos (3 x) d x))

Therefore

\int e^{4 x} cos (3 x) d x = \frac{1}{3} e^{4 x} sin (3 x) - \frac{4}{3} (- \frac{1}{3} e^{4 x} cos (3 x) - (- \frac{4}{3} \cdot \int e^{4 x} cos (3 x) d x))

Isolate

\int e^{4 x} cos (3 x) d x

= \frac{3 e ^{4 x} sin ( 3 x )}{25} + \frac{4 e ^{4 x} cos ( 3 x )}{25}

Add a constant to the solution

= \frac{3 e ^{4 x} sin ( 3 x )}{25} + \frac{4 e ^{4 x} cos ( 3 x )}{25} + C

\frac{\partial}{\partial x} (y + e^{x})

\int_{0}^{1} \frac{2 x}{1 + x ^{4}} d x

f (x) = \frac{- 4 x}{x ^{2} + 1}, at x = 0

x \to 3 lim (\frac{x ^{2} - 6}{x - 3})

\frac{d ^{2}}{d x ^{2}} (\frac{2}{3 t ^{4}})

What is the integral of e^{4x}cos(3x) ?
The integral of e^{4x}cos(3x) is (3e^{4x}sin(3x))/(25)+(4e^{4x}cos(3x))/(25)+C