Find the indefinite integral. (Use C for the constant of integrati

Ask by Pierce Bowman. in the United States

Mar 25,2025

Question

Find the indefinite integral. (Use C for the constant of integrati $$ \int \sin ^{4}(3 	heta) d 	heta $$

UpStudy AI · Accepted Answer

We start with
$$
\int \sin^4(3\theta) \, d\theta.
$$
**Step 1.** Rewrite the integrand using the identity
$$
\sin^2 x=\frac{1}{2}\left(1-\cos(2x)\right).
$$
Thus,
$$
\sin^4(3\theta)=\left[\sin^2(3\theta)\right]^2=\left[\frac{1}{2}\bigl(1-\cos(6\theta)\bigr)\right]^2=\frac{1}{4}\left(1-2\cos(6\theta)+\cos^2(6\theta)\right).
$$

**Step 2.** Express $$\cos^2(6\theta)$$ in terms of a cosine of a double angle:
$$
\cos^2(6\theta)=\frac{1}{2}\left(1+\cos(12\theta)\right).
$$
Substitute this into the expression for $$\sin^4(3\theta)$$:
$$
\sin^4(3\theta)=\frac{1}{4}\left(1-2\cos(6\theta)+\frac{1}{2}\bigl(1+\cos(12\theta)\bigr)\right).
$$
Simplify the expression by combining like terms:
$$
1-2\cos(6\theta)+\frac{1}{2}+\frac{1}{2}\cos(12\theta)
=\frac{3}{2}-2\cos(6\theta)+\frac{1}{2}\cos(12\theta).
$$
Thus,
$$
\sin^4(3\theta)=\frac{1}{4}\left(\frac{3}{2}-2\cos(6\theta)+\frac{1}{2}\cos(12\theta)\right).
$$

**Step 3.** Write the integral in its new form:
$$
\int \sin^4(3\theta) \, d\theta = \frac{1}{4}\int \left(\frac{3}{2}-2\cos(6\theta)+\frac{1}{2}\cos(12\theta)\right)d\theta.
$$

**Step 4.** Integrate term by term.

1. For the constant term:
   $$
   \int \frac{3}{2}\, d\theta = \frac{3}{2}\theta.
   $$
2. For the cosine term with argument $$6\theta$$:
   $$
   \int \cos(6\theta)\, d\theta = \frac{1}{6}\sin(6\theta).
   $$
   Multiplying by $$-2$$,
   $$
   -2\int \cos(6\theta)d\theta = -2\cdot\frac{1}{6}\sin(6\theta) = -\frac{1}{3}\sin(6\theta).
   $$
3. For the cosine term with argument $$12\theta$$:
   $$
   \int \cos(12\theta)\, d\theta = \frac{1}{12}\sin(12\theta).
   $$
   Multiplying by $$\frac{1}{2}$$,
   $$
   \frac{1}{2}\int \cos(12\theta)d\theta = \frac{1}{2}\cdot\frac{1}{12}\sin(12\theta) = \frac{1}{24}\sin(12\theta).
   $$

**Step 5.** Combine the results:
$$
\int \sin^4(3\theta) \, d\theta = \frac{1}{4}\left(\frac{3}{2}\theta -\frac{1}{3}\sin(6\theta) +\frac{1}{24}\sin(12\theta)\right)+C.
$$
Multiply through by $$\frac{1}{4}$$:
$$
\int \sin^4(3\theta) \, d\theta = \frac{3\theta}{8} -\frac{\sin(6\theta)}{12} +\frac{\sin(12\theta)}{96}+C.
$$

Thus, the final answer is:
$$
\boxed{\frac{3\theta}{8} -\frac{\sin(6\theta)}{12} +\frac{\sin(12\theta)}{96}+C.}
$$
The integral of $$ \sin^4(3\theta) $$ with respect to $$ \theta $$ is: $$ \frac{3\theta}{8} - \frac{\sin(6\theta)}{12} + \frac{\sin(12\theta)}{96} + C $$ where $$ C $$ is the constant of integration.

Find the indefinite integral. (Use C for the constant of integrati

Solución de inteligencia artificial de Upstudy

Responder

Solución

Mind Expander

preguntas relacionadas

Latest Calculus Questions