Integrate sin6xcos5xdx?

recall that

sinA cosB = [sin(A+B)+sin(A-B)]/2

so, your integrand is just

1/2(sin11x + sinx) dx

which is easy, right?

To integrate the function sin(6x)cos(5x)dx, you can use the product-to-sum identities and the power-reduction formula to simplify the expression.

Step 1: Apply the product-to-sum identities.

The product-to-sum identities state that sin(A)cos(B) = 0.5[sin(A + B) + sin(A - B)].

Using this identity, we can rewrite the integral as:

∫ sin(6x)cos(5x) dx = 0.5 ∫ [sin(6x+5x) + sin(6x-5x)] dx

= 0.5 ∫ [sin(11x) + sin(x)] dx

Step 2: Integrate term by term.

Since the integral of sin(x) is -cos(x), we can write:

0.5 ∫ [sin(11x) + sin(x)] dx = 0.5 [-cos(11x)/11 - cos(x)] + C

where C is the constant of integration.

Therefore, the integral of sin(6x)cos(5x)dx is:

∫ sin(6x)cos(5x) dx = 0.5 [-cos(11x)/11 - cos(x)] + C