1 Answer | Add Yours
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
We’ve answered 319,865 questions. We can answer yours, too.Ask a question