\( \leftarrow \int \) Evaluate the integral using any appropriate algebraic method or trigonometric identity. \( \int \frac{3 d x}{x \sqrt{4-9 \ln ^{2} x}} \)
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To evaluate the integral \( \int \frac{3 \, dx}{x \sqrt{4 - 9 \ln^2 x}} \), we can use a substitution. Let \( u = \ln x \), which gives \( du = \frac{1}{x} \, dx \) or \( dx = x \, du = e^u \, du \). Thus, the integral becomes: \[ \int \frac{3 \, e^u \, du}{e^u \sqrt{4 - 9u^2}} = 3 \int \frac{du}{\sqrt{4 - 9u^2}}. \] This integral resembles the integral of the form \( \int \frac{du}{\sqrt{a^2 - u^2}} \), which is solved using trigonometric substitution. Here, we have \( a^2 = 4 \) and \( 9u^2 = 4 - 9u^2 \) indicates a use of sine. We can set \( u = \frac{2}{3} \sin \theta \), which leads us to: \[ du = \frac{2}{3} \cos \theta \, d\theta. \] Transform the limits as necessary, and now integrate: \[ 3 \int \frac{\frac{2}{3} \cos \theta \, d\theta}{\sqrt{4 - 4 \sin^2 \theta}} = 3 \int \frac{\frac{2}{3} \cos \theta \, d\theta}{\sqrt{4(1 - \sin^2 \theta)}} = 3 \int \frac{\frac{2}{3} \cos \theta \, d\theta}{2 \cos \theta} = 3 \int \frac{1}{3} d\theta = \theta + C. \] Reverting back to \( u \) and then to \( x \), we find: \[ \theta = \arcsin\left(\frac{3}{2} u\right). \] Finally, substituting back, we have: \[ \int \frac{3 \, dx}{x \sqrt{4 - 9 \ln^2 x}} = \arcsin\left(\frac{3}{2} \ln x\right) + C. \]
