What is the integral of (sin(4x))e^x ?

The integral of (sin(4x))e^x is -(4e^xcos(4x))/(17)+(e^xsin(4x))/(17)+C

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integral of (sin(4x))e^x

\int (sin (4 x)) e^{x} d x

- \frac{4 e ^{x} cos ( 4 x )}{17} + \frac{e ^{x} sin ( 4 x )}{17} + C

Solution steps

\int sin (4 x) e^{x} d x

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

\int e^{x} sin (4 x) d x

= - \frac{4 e ^{x} cos ( 4 x )}{17} + \frac{e ^{x} sin ( 4 x )}{17}

Add a constant to the solution

= - \frac{4 e ^{x} cos ( 4 x )}{17} + \frac{e ^{x} sin ( 4 x )}{17} + C

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What is the integral of (sin(4x))e^x ?
The integral of (sin(4x))e^x is -(4e^xcos(4x))/(17)+(e^xsin(4x))/(17)+C