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Use a change of variables to evaluate the following definite integral. \[ \int_{\frac{\pi}{3}}^{\frac{5 \pi}{6}} \frac{\cos x}{\sin ^{2} x} d x \] What is the best choice of \( u \) for the change of variables? \[ \mathrm{u}=\square \] Find du. \[ d u=(\square d x \] Rewrite the given integral using this change of variables. \[ \int_{\frac{\pi}{3}}^{\frac{5 \pi}{6}} \frac{\cos x}{\sin ^{2} x} d x=\int_{\square}^{\square}(\square) d u \] (Type exact answers.)

Ask by Beck Ray. in the United States
Mar 08,2025

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Let \( u = \sin x \). Then, \( du = \cos x \, dx \). Rewritten integral: \[ \int_{\frac{\pi}{3}}^{\frac{5\pi}{6}} \frac{\cos x}{\sin^2 x} \, dx = \int_{\frac{\sqrt{3}}{2}}^{\frac{1}{2}} \frac{1}{u^2} \, du \]

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To effectively solve the integral, the best choice of \( u \) is \[ u = \sin x \] To find \( du \), we differentiate \( u \): \[ d u = \cos x \, d x \] Now, we need to find the new limits of integration. When \( x = \frac{\pi}{3} \), \[ u = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] And when \( x = \frac{5\pi}{6} \), \[ u = \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \] Next, we substitute \( u \) and \( du \) into the integral: \[ \int_{\frac{\pi}{3}}^{\frac{5 \pi}{6}} \frac{\cos x}{\sin ^{2} x} d x = \int_{\frac{\sqrt{3}}{2}}^{\frac{1}{2}} \frac{1}{u^2} \, du \] Thus, the rewritten integral is: \[ \int_{\frac{\sqrt{3}}{2}}^{\frac{1}{2}} \frac{1}{u^2} \, d u \]

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