What is the integral of sin(1-2x)cos(1+2x) ?

The integral of sin(1-2x)cos(1+2x) is 1/2 (sin(2)x+1/4 cos(4x))+C

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integral of sin(1-2x)cos(1+2x)

\int sin (1 - 2 x) cos (1 + 2 x) d x

\frac{1}{2} (sin (2) x + \frac{1}{4} cos (4 x)) + C

Solution steps

\int sin (1 - 2 x) cos (1 + 2 x) d x

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int \frac{sin ( 1 - 2 x + 1 + 2 x ) + sin ( 1 - 2 x - ( 1 + 2 x ) )}{2} 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 sin (1 - 2 x + 1 + 2 x) + sin (1 - 2 x - (1 + 2 x)) d x

Simplify

sin (1 - 2 x + 1 + 2 x) + sin (1 - 2 x - (1 + 2 x)) : sin (2) - sin (4 x)

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int sin (2) d x = sin (2) x

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

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

Simplify

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

Add a constant to the solution

= \frac{1}{2} (sin (2) x + \frac{1}{4} cos (4 x)) + C

x \to 0 lim (\frac{x ^{4} + 3 x}{x})

f (x) = x + x, at x = 9

x \to - 1 - lim (\frac{x}{1 + x ^{3}})

\frac{d}{d x} ((8 x^{2} - 5 x) (13 x^{2} + 4))

What is the integral of sin(1-2x)cos(1+2x) ?
The integral of sin(1-2x)cos(1+2x) is 1/2 (sin(2)x+1/4 cos(4x))+C