Questions: ∫ sin^3 3x dx

Transcript text: $\int \sin ^{3} 3 x d x$

Solution

Solution Steps

To solve the integral of $\sin^3(3x)$, we can use a trigonometric identity to simplify the expression. We know that $\sin^3(x) = \sin(x) \cdot \sin^2(x)$, and we can use the identity $\sin^2(x) = 1 - \cos^2(x)$ to rewrite the integral. Then, we can use substitution to solve the integral.

Step 1: Simplify the Integral Using Trigonometric Identities

To solve the integral $\int \sin^3(3x) \, dx$, we start by using the identity $\sin^2(x) = 1 - \cos^2(x)$. This allows us to rewrite $\sin^3(3x)$ as $\sin(3x) \cdot (1 - \cos^2(3x))$.

Step 2: Expand the Expression

Expanding the expression, we have: \[ \sin(3x) - \sin(3x) \cos^2(3x) \] This can be integrated term by term.

Step 3: Integrate Each Term

Integrate $\sin(3x)$: \[ \int \sin(3x) \, dx = -\frac{1}{3} \cos(3x) \]
Integrate $-\sin(3x) \cos^2(3x)$: Use substitution. Let $u = \cos(3x)$, then $du = -3 \sin(3x) \, dx$, or $\sin(3x) \, dx = -\frac{1}{3} du$. \[ \int -\sin(3x) \cos^2(3x) \, dx = \int u^2 \left(-\frac{1}{3}\right) \, du = -\frac{1}{3} \int u^2 \, du \] \[ = -\frac{1}{3} \cdot \frac{u^3}{3} = -\frac{1}{9} u^3 = -\frac{1}{9} \cos^3(3x) \]

Step 4: Combine the Results

Combine the results of the integrals: \[ -\frac{1}{3} \cos(3x) - \frac{1}{9} \cos^3(3x) \]