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

The integral of e^{2x}sin(x)cos(x) is 1/8 (-e^{2x}cos(2x)+e^{2x}sin(2x))+C

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of e^{2x}sin(x)cos(x) ?
The integral of e^{2x}sin(x)cos(x) is 1/8 (-e^{2x}cos(2x)+e^{2x}sin(2x))+C