91Ó°ÊÓ

First, use the product-to-sum identities to simplify the integrand: \(2 \sin 3x \cos 2x = \sin(3x + 2x) + \sin(3x - 2x) = \sin 5x + \sin x\). The integral then becomes:\[I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\sin 5x + \sin x) \mathrm{d}x\]

Calculate the integral of each term separately. The integral of \(\sin 5x\) is: \[\int \sin 5x \mathrm{d}x = -\frac{1}{5} \cos 5x\]The integral of \(\sin x\) is:\[\int \sin x \mathrm{d}x = -\cos x\]Now integrate both terms over the given limits.

Evaluate the definite integrals from \(\frac{\pi}{6} \) to \(\frac{\pi}{4}\) for both terms:\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \sin 5x \mathrm{d}x = -\frac{1}{5} \left[ \cos 5x \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = -\frac{1}{5} \left( \cos \left(\frac{5\pi}{4}\right) - \cos \left(\frac{5\pi}{6}\right) \right) \]\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \sin x \mathrm{d}x = - \left[ \cos x \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = - (\cos \left(\frac{\pi}{4}\right) - \cos \left(\frac{\pi}{6}\right)) \]

Determine the values of the trigonometric functions:\(\cos \left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}\), \(\cos \left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\), \(\cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), and \(\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\).Use these values to find:\[\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \sin 5x \mathrm{d} x = -\frac{1}{5} \left( -\frac{\sqrt{2}}{2} - \left( -\frac{\sqrt{3}}{2} \right) \right) = \frac{1}{5} \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \right) = \frac{1}{5} \left( \frac{\sqrt{3}-\sqrt{2}}{2} \right) \]\[\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \sin x \mathrm{d} x = - \left( \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \right) = - \left( \frac{\sqrt{2}-\sqrt{3}}{2} \right) = \frac{\sqrt{3}-\sqrt{2}}{2} \]

Add the results from the integrals:\[\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\sin 5x + \sin x) \mathrm{d} x = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \sin 5x \mathrm{d} x + \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \sin x \mathrm{d} x = \frac{\sqrt{3}-\sqrt{2}}{10} + \frac{\sqrt{3}-\sqrt{2}}{2} = 0.71\] (to two significant figures)

Use the substitution \(t = \tan x\), \(\cos x = \frac{1}{\sqrt{1+t^2}}\), and \(\sin x = \frac{t}{\sqrt{1+t^2}}\). The differential \(\mathrm{d}x = \frac{1}{1+t^2} \mathrm{d}t\). Substitute into the integral:\[\int \frac{\mathrm{d}x}{4 \cos^2 x - 9 \sin^2 x} = \int \frac{\frac{1}{1+t^2}\mathrm{d}t}{4\left(\frac{1}{1+t^2}\right) - 9\left(\frac{t^2}{1+t^2}\right)} =\int \frac{\mathrm{d}t}{4 - 9t^2}\]

Perform partial fraction decomposition:\[\int \frac{\mathrm{d}t}{4-9t^2} = \int \frac{\mathrm{d}t}{(2+3t)(2-3t)} = \int \left( \frac{A}{2+3t} + \frac{B}{2-3t} \right) \mathrm{d}t\]Solve for A and B:\[1 = A(2-3t) + B(2+3t)\] and solve for A and B.

After solving for A and B, integrate each term independently:\[\int \frac{\mathrm{d}t}{4-9t^2} = \int \left( \frac{1}{6(2+3t)} - \frac{1}{6(2-3t)} \right) \mathrm{d}t \]\[= \frac{1}{6} \left( \int \frac{\mathrm{d}t}{2+3t} - \int \frac{\mathrm{d}t}{2-3t} \right) = \frac{1}{6} \left( \frac{1}{3} \ln |2+3t| + \frac{1}{3} \ln |2-3t| \right)\]

Replace \(t\) back with \(\tan x\). The result is:\[\frac{1}{9} \ln \left| \frac{2+3\tan x}{2-3\tan x} \right| + C\]