Integrate $\int \frac{\sin(3x)}{1-\cos^2(\frac{3x}{2})}dx$

Question

Integrate $$\int \frac{\sin(3x)}{1-\cos^2(\frac{3x}{2})}dx$$

My attempt,

Let $u=1-\cos^2(\frac{3x}{2})$. Then

$$du=-2\cdot\cos\left(\frac{3x}{2}\right)\cdot -\sin\left(\frac{3x}{2}\right)\cdot \frac{3}{2}=3\sin\left(\frac{3x}{2}\right)\cos\left(\frac{3x}{2}\right)\,dx$$

How should I continue? Thanks in advance.

Answer 1

$$\int \frac{\sin3x}{1-\cos^2(\frac{3x}{2})}\,dx=\int\frac{2\sin\frac{3x}2\cos\frac{3x}2}{\sin^2(\frac{3x}{2})}\,dx=2\int \cot\frac{3x}2\,dx=\frac43\ln\bigg|\sin\frac{3x}2\bigg|+C$$

Answer 2

By double-angle formula, the numerator is

$$\sin 3x = 2\sin(3x/2)\cos(3x/2).$$

This is pretty close to your expression for $du$.

Answer 3

Recall that $$\cos^2(x)=\frac{1+\cos(2x)}{2}\implies 1-\cos^2\left(\frac{3x}{2}\right)=1-\frac{1+\cos(3x)}{2}=\frac{1-\cos(3x)}{2}$$ Then you could try a substitution like $u=\cos\left(3x\right)$.

Another way to do this: notice that $1-\cos^2\left(\frac{3x}{2}\right)=\sin^2\left(\frac{3x}{2}\right)$ and $\sin(3x)=\frac{\sin\left(\frac{3x}{2}\right)\cos\left(\frac{3x}{2}\right)}{2}$, and then much in the integrand cancels out.