What is the integral of 4sin(x+pi/4)cos(x-pi/4) ?

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

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integral of 4sin(x+pi/4)cos(x-pi/4)

\int 4 sin (x + \frac{π}{4}) cos (x - \frac{π}{4}) d x

- cos (2 x) + 2 x + C

Solution steps

\int 4 sin (x + \frac{π}{4}) cos (x - \frac{π}{4}) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int sin (x + \frac{π}{4}) cos (x - \frac{π}{4}) d x

Use the following identity:

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Simplify

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

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

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

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

\int 1 d x = x

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

Simplify

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

= - cos (2 x) + 2 x

Add a constant to the solution

= - cos (2 x) + 2 x + C

\frac{\partial}{\partial x} (\frac{x ( v - b )}{r ( x , v )})

\int \frac{sin ( x ) \cdot cos ( x )}{x} d x

\int \frac{1}{- 196.2 - 4 v ^{2}} d v

x \to \infty lim (\frac{4}{x})

\frac{d}{d x} ((sec (x))^{3})

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