# Integrate the function sin3x*cos5xIntegrate the function sin3x*cos5x

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You also may use the following alternative method, such that:

`sin 3x = sin (5x - 2x) => sin 3x = sin 5x*cos 2x - sin 2x* cos 5x`

`sin 3x*cos 5x = (sin 5x*cos 2x - sin 2x* cos 5x)*cos 5x`

`sin 3x*cos 5x = sin 5x*cos 5x*cos 2x - sin 2x*cos^2 5x`

Using the double angle identity yields:

`sin 5x*cos 5x = (sin 2*(5x))/2 => sin 5x*cos 5x = (sin 10 x)/2`

Using the half angle identity yields:

`cos^2 5x = (1 + cos 10 x)/2`

`sin 3x*cos 5x = (sin 10 x)/2*cos 2x - sin 2x*(1 + cos 10 x)/2`

`sin 3x*cos 5x = ((sin 10 x)*(cos 2x) - (sin 2x)*(cos 10 x) - sin 2x)/2`

You need to replace `sin (10x - 2x)` for `(sin 10 x)*(cos 2x) - (sin 2x)*(cos 10 x),` such that:

`sin 3x*cos 5x = (sin (10x - 2x) - sin 2x)/2`

You need to integrate the identity above, such that:

`int sin 3x*cos 5x dx = int (sin (10x - 2x) - sin 2x)/2 dx`

Using the property of linearity of integral yields:

`int sin 3x*cos 5x dx = int (sin 8x)/2 dx - int (sin 2x)/2 dx`

`int sin 3x*cos 5x dx = -(cos 8x)/16 + (cos 2x)/4 + c`

**Hence, evaluating the given integral using trigonometric identities, yields **`int sin 3x*cos 5x dx = (1/4)(cos 2x - (cos 8x)/4) + c.`

To calculate the indefinite integral of f(x)=sin3x*cos5x, we'll transform the product of trigonometric functions into a sum.

We'll use the formula:

sin a * cos b = [sin(a+b)+sin(a-b)]/2

We'll substitute a by 3x and b by 5x.

sin3x*cos5x = [sin(3x+5x)+sin(3x-5x)]/2

sin3x*cos5x = (sin 8x)/2 - (sin 2x)/2

Now, we'll calculate Int f(x)dx.

Int sin3x*cos5x dx = Int (sin 8x)dx/2 - Int (sin 2x)dx/2

Int (sin 8x)dx = -(cos8x)/8 + C

Int (sin2x)dx = -(cos 2x)/2 + C

Int sin3x*cos5x dx = -(cos8x)/16 + (cos 2x)/4 + C