What is the integral of e^{4x}cos(-1x) ?

The integral of e^{4x}cos(-1x) is -16/17 e^{4x}sin(x)+4/17 e^{4x}cos(x)+e^{4x}sin(x)+C

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integral of e^{4x}cos(-1x)

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

- \frac{16}{17} e^{4 x} sin (x) + \frac{4}{17} e^{4 x} cos (x) + e^{4 x} sin (x) + C

Solution steps

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

= - (\frac{16 e ^{4 x} sin ( x )}{17} - \frac{4 e ^{4 x} cos ( x )}{17}) + sin (x) e^{4 x}

Simplify

- (\frac{16 e ^{4 x} sin ( x )}{17} - \frac{4 e ^{4 x} cos ( x )}{17}) + sin (x) e^{4 x} : - \frac{16}{17} e^{4 x} sin (x) + \frac{4}{17} e^{4 x} cos (x) + e^{4 x} sin (x)

= - \frac{16}{17} e^{4 x} sin (x) + \frac{4}{17} e^{4 x} cos (x) + e^{4 x} sin (x)

Add a constant to the solution

= - \frac{16}{17} e^{4 x} sin (x) + \frac{4}{17} e^{4 x} cos (x) + e^{4 x} sin (x) + C

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What is the integral of e^{4x}cos(-1x) ?
The integral of e^{4x}cos(-1x) is -16/17 e^{4x}sin(x)+4/17 e^{4x}cos(x)+e^{4x}sin(x)+C