How do you find the integral of the function sin(x) cos(x) plus 3sin^2(x) with respect to x?

To solve the integral of the function sin(x) cos(x) + 3sin^2(x), we can break it down into two parts: sin(x) cos(x) and 3sin^2(x).

The integral of sin(x) cos(x) can be simplified using a trigonometric identity:

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

Therefore, we have:

∫sin(x) cos(x) dx = ∫(1/2) sin(2x) dx

Now, we can integrate sin(2x):

∫sin(2x) dx = -1/2 cos(2x) + C

So, the integral of sin(x) cos(x) is:

(1/2)(-1/2 cos(2x)) + C = -1/4 cos(2x) + C

For the integral of 3sin^2(x), we can use the power-reduction formula:

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

This gives us:

3sin^2(x) = (3/2)(1 - cos(2x))

Now we integrate:

∫(3/2)(1 - cos(2x)) dx = (3/2) ∫(1 - cos(2x)) dx

This can be separated into two integrals:

(3/2)(x - (1/2)sin(2x)) + C

So, the integral of 3sin^2(x) is:

(3/2)x - (3/4)sin(2x) + C

Now, we combine the results of both integrals. Therefore, the complete integral is:

-1/4 cos(2x) + (3/2)x - (3/4)sin(2x) + C

So, the integral of sin(x) cos(x) + 3sin^2(x) is:

∫(sin(x) cos(x) + 3sin^2(x)) dx = -1/4 cos(2x) + (3/2)x – (3/4)sin(2x) + C