$$ \int \sin ^{2} 4 x d x $$ is equal to

Question

UpStudy AI · Accepted Answer

To evaluate the integral $$ \int \sin^{2}(4x) \, dx $$, we can use the trigonometric identity for $$ \sin^{2}(\theta) $$:

$$
\sin^{2}(\theta) = \frac{1 - \cos(2\theta)}{2}
$$

Applying this identity to $$ \sin^{2}(4x) $$:

$$
\sin^{2}(4x) = \frac{1 - \cos(8x)}{2}
$$

Now, substitute this back into the integral:

$$
\int \sin^{2}(4x) \, dx = \int \frac{1 - \cos(8x)}{2} \, dx = \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos(8x) \, dx
$$

Integrate term by term:

$$
\frac{1}{2} \int 1 \, dx = \frac{1}{2} x
$$

$$
-\frac{1}{2} \int \cos(8x) \, dx = -\frac{1}{2} \cdot \frac{\sin(8x)}{8} = -\frac{\sin(8x)}{16}
$$

Combine the results and add the constant of integration $$ C $$:

$$
\int \sin^{2}(4x) \, dx = \frac{1}{2} x - \frac{\sin(8x)}{16} + C
$$

**Final Answer:**

$$
\int \sin^{2}(4x) \, dx = \frac{1}{2} x - \frac{\sin(8x)}{16} + C
$$
$$ \int \sin^{2}(4x) \, dx = \frac{1}{2} x - \frac{\sin(8x)}{16} + C $$

is equal to