What is the integral of 4e^xcos(2x) ?

The integral of 4e^xcos(2x) is 4((2e^xsin(2x))/5+(e^xcos(2x))/5)+C

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integral of 4e^xcos(2x)

\int 4 e^{x} cos (2 x) d x

4 (\frac{2 e ^{x} sin ( 2 x )}{5} + \frac{e ^{x} cos ( 2 x )}{5}) + C

Solution steps

\int 4 e^{x} cos (2 x) d x

Take the constant out:

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

= 4 \cdot \int e^{x} cos (2 x) d x

Apply Integration By Parts

: e^{x} \frac{1}{2} sin (2 x) - \int e^{x} \frac{1}{2} sin (2 x) d x

= 4 (e^{x} \frac{1}{2} sin (2 x) - \int e^{x} \frac{1}{2} sin (2 x) d x)

Take the constant out:

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

= 4 (e^{x} \frac{1}{2} sin (2 x) - \frac{1}{2} \cdot \int e^{x} sin (2 x) d x)

Apply Integration By Parts

: - \frac{1}{2} e^{x} cos (2 x) - \int - \frac{1}{2} e^{x} cos (2 x) d x

= 4 (e^{x} \frac{1}{2} sin (2 x) - \frac{1}{2} (- \frac{1}{2} e^{x} cos (2 x) - \int - \frac{1}{2} e^{x} cos (2 x) d x))

Take the constant out:

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

= 4 (e^{x} \frac{1}{2} sin (2 x) - \frac{1}{2} (- \frac{1}{2} e^{x} cos (2 x) - (- \frac{1}{2} \cdot \int e^{x} cos (2 x) d x)))

Therefore

4 \cdot \int e^{x} cos (2 x) d x = 4 (e^{x} \frac{1}{2} sin (2 x) - \frac{1}{2} (- \frac{1}{2} e^{x} cos (2 x) - (- \frac{1}{2} \cdot \int e^{x} cos (2 x) d x)))

Isolate

\int e^{x} cos (2 x) d x

= 4 (\frac{2 e ^{x} sin ( 2 x )}{5} + \frac{e ^{x} cos ( 2 x )}{5})

Add a constant to the solution

= 4 (\frac{2 e ^{x} sin ( 2 x )}{5} + \frac{e ^{x} cos ( 2 x )}{5}) + C

f (x) = (1 + 3 x) (5 - 2 x), at x = 2

\frac{s}{( s + 1 ) ( s ^{2} + 4 )}

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x ^{10}})

\int - tan (t) d t

\frac{3}{( s + 1 ) ^{2} + 1}

What is the integral of 4e^xcos(2x) ?
The integral of 4e^xcos(2x) is 4((2e^xsin(2x))/5+(e^xcos(2x))/5)+C