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

The integral of cos(x)e^{4x} is (e^{4x}sin(x))/(17)+(4e^{4x}cos(x))/(17)+C

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

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What is the integral of cos(x)e^{4x} ?
The integral of cos(x)e^{4x} is (e^{4x}sin(x))/(17)+(4e^{4x}cos(x))/(17)+C