What is 3/8 (2x^2sin(2x)+2xcos(2x)-sin(2x)) ?

The solution to 3/8 (2x^2sin(2x)+2xcos(2x)-sin(2x)) is (3x^2sin(2x)+3xcos(2x))/4-(3sin(2x))/8

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simplify 3/8 (2x^2sin(2x)+2xcos(2x)-sin(2x))

simplify

\frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

\frac{3 x ^{2} sin ( 2 x ) + 3 x cos ( 2 x )}{4} - \frac{3 sin ( 2 x )}{8}

Solution steps

\frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

Expand

\frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)) : \frac{3 x ^{2} sin ( 2 x ) + 3 x cos ( 2 x )}{4} - \frac{3 sin ( 2 x )}{8}

= \frac{3 x ^{2} sin ( 2 x ) + 3 x cos ( 2 x )}{4} - \frac{3 sin ( 2 x )}{8}

\frac{3 ^{3^{3}} \cdot 3 ^{3} \cdot 3}{( 3 ^{3} ) ^{9}}

2 (2 + \times (- (6 - 7 x)) = 13 x - (1 + 4 x))

(\frac{x}{y} - y) (\frac{x}{y} + y)

(x - 2 - 6) \cdot (x - 2 + 6)

(26) l e f t (x^{4} - 81)

What is 3/8 (2x^2sin(2x)+2xcos(2x)-sin(2x)) ?
The solution to 3/8 (2x^2sin(2x)+2xcos(2x)-sin(2x)) is (3x^2sin(2x)+3xcos(2x))/4-(3sin(2x))/8