# Which is the indefinite integral of f(x)=sin5x*cos3x ?

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To calculate the indefinite integral of f(x)=sin5x*cos3x, we'll apply the formula to 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 5x and b by 3x.

sin5x*cos3x = [sin(5x+3x)+sin(5x-3x)]/2

sin5x*cos3x = sin 8x/2 + sin2x/2

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

Int sin5x*cos3xdx = Int (sin 8x)dx/2 + Int (sin2x)dx/2

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

Int (sin2x)dx = -(cos2x)/2 + C

**Int sin5x*cos3xdx = -(cos8x)/8 - (cos2x)/2 + C**

To find the integgral of sin5x*cos3x.

We know sinA*cosB = (1/2){sin (A+B) + Cos(A-B)}

Therefore sin5x*cos3x = (1/2) sin (5+3)x +(1/2) sin (5-3x)

Therefore Integral sin5x*cos3x = (1/2) Intgral {sin8x+sin2x}dx

= (1/2) {(-cos8x)/8-(cos2x)/2 = -(1/4){(cos8x)/4 + sin2x} + C , where C is the constant of integration.