What is expand (e^xsin(2x))(e^xcos(2x)-2e^xsin(2x)) ?

The solution to expand (e^xsin(2x))(e^xcos(2x)-2e^xsin(2x)) is e^{2x}cos(2x)sin(2x)-2e^{2x}sin^2(2x)

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expand (e^xsin(2x))(e^xcos(2x)-2e^xsin(2x))

expand

(e^{x} sin (2 x)) (e^{x} cos (2 x) - 2 e^{x} sin (2 x))

e^{2 x} cos (2 x) sin (2 x) - 2 e^{2 x} sin^{2} (2 x)

Solution steps

(e^{x} sin (2 x)) (e^{x} cos (2 x) - 2 e^{x} sin (2 x))

Apply the distributive law:

a (b - c) = ab - a c

a = (e^{x} sin (2 x)), b = e^{x} cos (2 x), c = 2 e^{x} sin (2 x)

= (e^{x} sin (2 x)) e^{x} cos (2 x) - (e^{x} sin (2 x)) \cdot 2 e^{x} sin (2 x)

= e^{x} cos (2 x) (e^{x} sin (2 x)) - 2 e^{x} sin (2 x) (e^{x} sin (2 x))

Simplify

e^{x} cos (2 x) (e^{x} sin (2 x)) - 2 e^{x} sin (2 x) (e^{x} sin (2 x)) : e^{2 x} cos (2 x) sin (2 x) - 2 e^{2 x} sin^{2} (2 x)

= e^{2 x} cos (2 x) sin (2 x) - 2 e^{2 x} sin^{2} (2 x)

- 2 (4 e^{- x^{2}} x^{3} - 6 e^{- x^{2}} x)

\frac{x + 1}{2 ( x - 1 ) ( 4 x + 1 )}

(x + 5) l e f t (x - 5)

(r + sin (θ) - cos (θ)) d r

\frac{( s ^{3} + 2 )}{s ^{2} \cdot ( s ^{2} + 4 )}

What is expand (e^xsin(2x))(e^xcos(2x)-2e^xsin(2x)) ?
The solution to expand (e^xsin(2x))(e^xcos(2x)-2e^xsin(2x)) is e^{2x}cos(2x)sin(2x)-2e^{2x}sin^2(2x)