How do you find the integral of #int [sin(3x)]^4 dx#?

Question

How do you find the integral of #int   [sin(3x)]^4 dx#?

Evelyn Agard · Accepted Answer

Use power reduction formulas twice.

#cos(2x) = cos^2x-sin^2x# #cos(2x) = 1-2sin^2x# #cos(2x) = 2cos^2x-1# So, #sin^2x = 1/2(1-cos(2x))# and, #cos^2x = 1/2(1+cos(2x))# #[sin(3x)]^4 = [color(blue)(sin^2 3x)]^2# # = [color(blue)(1/2(1-cos6x))]^2# # = 1/4[1-2cos6x+color(red)(cos^2 6x)]# # = 1/4[1-2cos6x+(color(red)(1/2(1+cos12x)))]# # = 1/8[2(1-2cos6x)+(1+cos12x))]# # = 1/8[3-4cos6x+cos12x]# By substituting the two cosine-related terms, the final expression can be integrated term by term.

Christopher Agresta · Answer

undefined To find the integral of $\int (\sin(3x))^4 \, dx$, you can use the power-reducing identity for sine raised to an even power. The identity states:

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

Using this identity, you can express $(\sin(3x))^4$ as a function of $\cos(6x)$. Then integrate $\cos(6x)$ with respect to $x$ and apply the appropriate constants.

The integral of $\cos(6x)$ is $\frac{1}{6} \sin(6x) + C$, where $C$ is the constant of integration.

So, the integral of $(\sin(3x))^4$ with respect to $x$ is $\frac{1}{6} \sin(6x) + C$.

Axel Alvarez · Answer

undefined To find the integral of $ \int \sin^4(3x) \, dx $, you can use the power-reducing identity for sine raised to an even power. This identity states that $ \sin^2(x) = \frac{1 - \cos(2x)}{2} $. Using this identity twice, you can reduce $ \sin^4(3x) $ to a polynomial in terms of $ \cos(6x) $. Then integrate the resulting expression.