Find the indefinite integral. $$\int \sin ^ { 2 } 3 x d x$$
A) $\frac { 3 x - \sin 3 x \cos 3 x } { 3 } + C$
B) $\frac { 3 x + \sin 3 x \cos 3 x } { 6 } + C$
C) $\frac { 3 x - \sin 3 x \cos 3 x } { 6 } + C$
D) $\frac { 3 x - \sin ^ { 2 } 3 x \cos 3 x } { 6 } + C$
E) $\frac { 3 x + \sin 3 x \cos 3 x } { 3 } + C$

Question

Quizplus · Accepted Answer

The integral of $\sin^2(3x)$ can be found using the identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$. Applying this identity and integrating, we get $\frac{3x}{2} - \frac{\sin(6x)}{12} + C$, which simplifies to the given option C.

Find the Indefinite Integral ∫sin⁡23xdx\int \sin ^ { 2 } 3 x d x∫sin23xdx

Find the Indefinite Integral $\int \sin ^ { 2 } 3 x d x$