Find the integral of sin(x) cos(x) 3sin(2x) dx

To solve the integral of sin(x) cos(x) 3sin(2x) dx, we start by simplifying the expression. We know that sin(2x) = 2sin(x)cos(x), thus 3sin(2x) = 6sin(x)cos(x).

This means we can rewrite the integral as:

∫ sin(x) cos(x) 6sin(x) cos(x) dx

Combining terms, we have:

∫ 6sin^2(x) cos^2(x) dx

Next, we can use the identity for sin^2(x):

sin^2(x) = (1 - cos(2x)) / 2

Substituting into the integral gives:

∫ 6 * (1 - cos(2x))/2 * cos^2(x) dx

Thus:

∫ 3(1 - cos(2x)) cos^2(x) dx

Now, we can distribute:

∫ 3cos^2(x) dx - ∫ 3cos^2(x) cos(2x) dx

Each of these parts can be solved separately. The first part can be computed using the substitution:

cos^2(x) = (1 + cos(2x))/2

This results in:

∫ 3(1 + cos(2x))/2 dx = (3/2)x + (3/4)sin(2x)

For the second term, we can use integration by parts or further identities, but it’s more complex so you may use numerical methods or tables for simplification on that part.

In conclusion, the integral of sin(x) cos(x) 3sin(2x) can lead to more manageable parts that can be evaluated individually. The final solution involves combining these evaluated parts along with any constant of integration which should be added when computing indefinite integrals.