# Integrate the function sin 3x*cos5x.

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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**

intg sin3x * cos5x

Let us rewrite:

we know that:

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

==> sin3x *cos5x = [sin4x + sin2x]/2

= (1/2)sin4x + (1/2)sin2x

Now let us integrate:

==> intg sin3x*cos5x = (1/2) intg (sin4x + sin2x) dx

= (1/2) [ intg sin4x dx+ intg sin2x dx]

= (1/2) [(-cos4x)/4 + (-cos2x)/2] + C

** = -cos4x / 8 - cos2x/4 + C**

To integrate sin3x*cos5x .

We know that Int {u(x)v(x)} = u(x)Int v(x)dx - Int {u'(x) intv(x) dx}dx.

Let I = Int (sin3x *cos5x) dx = sin3x (sin5x)/5 - (1/5) Int ((sin3x )'*sin5x)dx

I = (1/5) sin3x sin5x - (1/5) Int (3cos3x*sin5x) dx

I = (1/5)sin3xsin5x - (3/5){cos3x* (-cos5x)/5 - Int (-3sin3x)(-cos5x/5)}dx

I = (1/5) sin3xsin5x +(3/25)(cos3x+cos5x) + (9/25) I

(1-9/25)I = (1/5)sin3xsin5x +(3/25)(cos3x+cos5x)

Divide both sides by (1-9/25) = 16/25:

I = (5/16)(sin3xsin5x) +(3/16)(cos3x+cos5x).

Therefore Int (sin3x*cos3x)dx = (5/16)(sin3xsin5x) +(3/16)(cos3x+cos5x).