# What is the integral of y=sin2x*cos6x ?

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To evaluate the indefinite integral of f(x)=sin2x*cos6x, 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 2x and b by 6x.

sin5x*cos3x = [sin(2x+6x)+sin(2x-6x)]/2

sin5x*cos3x = (sin 8x)/2 + [sin(-4x)]/2

Since the sine function is odd, we'll have [sin(-4x)]/2=-(sin4x)/2

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

Int sin2x*cos6xdx = (1/2)[Int (sin 8x)dx - Int (sin4x)dx]

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

Int (sin2x)dx = -(cos4x)/4 + C

**Int sin2x*cos6xdx = -(cos8x)/16 + (cos4x)/8 + C**