Questions: ∫ (cos 3x) / √(1-2 sen 3x) dx

Solution Steps

To solve the integral \(\int \frac{\cos 3x}{\sqrt{1-2 \sin 3x}} \, dx\), we can use a substitution method. Let \( u = 1 - 2 \sin 3x \), which simplifies the expression under the square root. Then, find \( du \) in terms of \( dx \) and substitute back into the integral. This will transform the integral into a form that is easier to solve.

We are given the integral \(\int \frac{\cos 3x}{\sqrt{1-2 \sin 3x}} \, dx\).

To simplify the integral, we use the substitution \( u = 1 - 2 \sin 3x \). This implies that \( du = -6 \cos 3x \, dx \), or equivalently, \( dx = -\frac{1}{6} \frac{du}{\cos 3x} \).

Substituting into the integral, we have: \[ \int \frac{\cos 3x}{\sqrt{u}} \left(-\frac{1}{6} \frac{du}{\cos 3x}\right) = -\frac{1}{6} \int \frac{1}{\sqrt{u}} \, du \]

The integral \(-\frac{1}{6} \int \frac{1}{\sqrt{u}} \, du\) is straightforward: \[ -\frac{1}{6} \cdot 2\sqrt{u} = -\frac{1}{3} \sqrt{u} \]

Substitute back \( u = 1 - 2 \sin 3x \) to get: \[ -\frac{1}{3} \sqrt{1 - 2 \sin 3x} \]

Final Answer