Consider $$ \int \sin ^{4}(a \boldsymbol{x}) d x $$ Expansion of $$ \sin ^{4}(a x) $$ gives us $$ \begin{array}{l}(3 / 8)-(1 / 2) \cos a x+(1 / 8) \cos 4 a x \ (1 / 8)+(1 / 2) \cos a x+(1 / 8) \cos 4 a x \ (1 / 8)+(1 / 2) \cos a x-(1 / 8) \cos 4 a x \ (3 / 8)+(1 / 2) \cos a x+(1 / 8) \cos 4 a x \ (1 / 8)+(3 / 2) \cos a x+(1 / 8) \cos 4 a x\end{array} $$

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To find the integral $$ \int \sin^4(a x) \, dx $$, we can use the trigonometric identity for $$ \sin^2(x) $$:

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

Using this identity, we can express $$ \sin^4(a x) $$ as follows:

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

Expanding this gives:

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

Next, we can use the identity for $$ \cos^2(x) $$:

$$
\cos^2(x) = \frac{1 + \cos(2x)}{2}
$$

Applying this to $$ \cos^2(2 a x) $$:

$$
\cos^2(2 a x) = \frac{1 + \cos(4 a x)}{2}
$$

Substituting this back into our expression for $$ \sin^4(a x) $$:

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

Simplifying this:

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

$$
= \frac{\frac{3}{2} - 2\cos(2 a x) + \frac{1}{2}\cos(4 a x)}{4}
$$

$$
= \frac{3}{8} - \frac{1}{2}\cos(2 a x) + \frac{1}{8}\cos(4 a x)
$$

Thus, we have:

$$
\sin^4(a x) = \frac{3}{8} - \frac{1}{2}\cos(2 a x) + \frac{1}{8}\cos(4 a x)
$$

Now, we can integrate $$ \sin^4(a x) $$:

$$
\int \sin^4(a x) \, dx = \int \left( \frac{3}{8} - \frac{1}{2}\cos(2 a x) + \frac{1}{8}\cos(4 a x) \right) dx
$$

Calculating each term separately:

1. $$ \int \frac{3}{8} \, dx = \frac{3}{8} x $$
2. $$ \int -\frac{1}{2} \cos(2 a x) \, dx = -\frac{1}{2} \cdot \frac{1}{2a} \sin(2 a x) = -\frac{1}{4a} \sin(2 a x) $$
3. $$ \int \frac{1}{8} \cos(4 a x) \, dx = \frac{1}{8} \cdot \frac{1}{4a} \sin(4 a x) = \frac{1}{32a} \sin(4 a x) $$

Combining these results, we have:

$$
\int \sin^4(a x) \, dx = \frac{3}{8} x - \frac{1}{4a} \sin(2 a x) + \frac{1}{32a} \sin(4 a x) + C
$$

where $$ C $$ is the constant of integration.

Thus, the final result is:

$$
\int \sin^4(a x) \, dx = \frac{3}{8} x - \frac{1}{4a} \sin(2 a x) + \frac{1}{32a} \sin(4 a x) + C
$$
The integral of $$ \sin^4(a x) $$ is $$ \frac{3}{8} x - \frac{1}{4a} \sin(2 a x) + \frac{1}{32a} \sin(4 a x) + C $$.

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