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

The integral of sin(x)cos(x)e^x is -(e^xcos(2x))/5+(e^xsin(2x))/(10)+C

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integral of sin(x)cos(x)e^x

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

- \frac{e ^{x} cos ( 2 x )}{5} + \frac{e ^{x} sin ( 2 x )}{10} + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{2} e^{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^{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

= \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

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

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

Therefore

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

Isolate

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

= - \frac{e ^{x} cos ( 2 x )}{5} + \frac{e ^{x} sin ( 2 x )}{10}

Add a constant to the solution

= - \frac{e ^{x} cos ( 2 x )}{5} + \frac{e ^{x} sin ( 2 x )}{10} + C

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\int \frac{1}{( 9 x ^{2} - 1 ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (\frac{1 - 4 x}{6 + x})

\int \frac{1}{( u ^{2} + 9 ) ^{\frac{5}{2}}} d u

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What is the integral of sin(x)cos(x)e^x ?
The integral of sin(x)cos(x)e^x is -(e^xcos(2x))/5+(e^xsin(2x))/(10)+C