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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

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6 e^{- 5 t} + e^{3 t} + 5 t^{3} - 9

\frac{\partial}{\partial u} (u - v)

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