How do you integrate (sin(2x+3))^2 ?

∫ ( sin(2x + 3) )^2 dx

knowing the fact :

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

so, we can rewrite as:

∫ sin^2(2x + 3) dx

∫ (1/2) * ( 1 - cos(2 * (2x + 3) ) ) dx

∫ (1/2) * ( 1 - cos(4x + 6) ) dx

(1/2) * ( x - (1/4)sin(4x + 6) ) + C

(1/2)x - (1/8)sin(4x + 6) + C

Hope it Helps :)

∫sin^2(2x + 3) dx

Let z = 2x + 3 => dz/2 = dx

1/2 * ∫sin^2(z) dz

Now note that sin^2(z) = 1/2 * (1 - cos(2z)), so:

1/4 * ∫(1 - cos(2z)) dz

1/4 * (z - 1/2*sin(2z)) + C

Note that sin(2z) = 2sin(z)cos(z), so:

1/4 * (z - sin(z)cos(z)) + C

Plug z back in:

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

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

Note that the 3/4 can be absorbed into the +C since it's a constant so the answer is:

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