# Indicate one method to solve the integral of y=sin4x*cos6x?

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### 1 Answer

The first step in evaluating the indefinite integral is to transform the given product of trigonometric functions into a sum.

We'll apply the identity:

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

We'll substitute a by 4x and b by 6x.

sin4x*cos6x = [sin(4x+6x)+sin(4x-6x)]/2

sin4x*cos6x = (sin 10x)/2 - (sin 2x)/2

Now, we'll evaluate the integral:

Int sin4x*cos6x dx = Int (sin 10x)dx/2 - Int (sin 2x)dx/2

Int (sin 10x)dx = -(cos 10x)/10 + C

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

**The indefinite integral of the given trigonometric product is: Int sin4x*cos6x dx = -(cos 10x)/20 + (cos 2x)/4 + C**