What is the integral of e^{4x}sin(2x)cos(2x) ?

The integral of e^{4x}sin(2x)cos(2x) is 1/16 (-e^{4x}cos(4x)+e^{4x}sin(4x))+C

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integral of e^{4x}sin(2x)cos(2x)

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

\frac{1}{16} (- e^{4 x} cos (4 x) + e^{4 x} sin (4 x)) + C

Solution steps

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

Rewrite using trig identities

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

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

= \frac{1}{2} (- \frac{e ^{4 x} cos ( 4 x )}{8} + \frac{e ^{4 x} sin ( 4 x )}{8})

Simplify

\frac{1}{2} (- \frac{e ^{4 x} cos ( 4 x )}{8} + \frac{e ^{4 x} sin ( 4 x )}{8}) : \frac{1}{16} (- e^{4 x} cos (4 x) + e^{4 x} sin (4 x))

= \frac{1}{16} (- e^{4 x} cos (4 x) + e^{4 x} sin (4 x))

Add a constant to the solution

= \frac{1}{16} (- e^{4 x} cos (4 x) + e^{4 x} sin (4 x)) + C

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What is the integral of e^{4x}sin(2x)cos(2x) ?
The integral of e^{4x}sin(2x)cos(2x) is 1/16 (-e^{4x}cos(4x)+e^{4x}sin(4x))+C