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Question Number 186840 by cortano12 last updated on 11/Feb/23

$$\underset{9}{\stackrel{16}{\int }}\frac{\sqrt{4-\sqrt{x}}}{x}dx=?$$

Answered by horsebrand11 last updated on 11/Feb/23

$$I=\underset{9}{\stackrel{16}{\int }}\frac{\sqrt{4-\sqrt{x}}}{x}dx=?$$$$let\sqrt{x}=4\sin {}^{2}t\Rightarrow x=16\sin {}^{4}t$$$$I=\int \frac{\sqrt{4-4\sin {}^{2}t}}{16\sin {}^{4}t}\left(64\sin {}^{3}t\cos t\right)dt$$$$=\int \frac{{8\cos }^{2}tdt}{\sin t}=8\int \csc tdt-8\int \sin tdt$$$$=8\ln \mid \csc t-\cot t\mid +8\cos t+C$$$$=8\ln \mid \frac{1-\cos t}{\sin t}\mid +8\cos t+C$$$$I=8{[\ln \mid \frac{1-\cos t}{\sin t}\mid +\cos t]}_{\pi /3}^{\pi /2}$$$$I=8[0-(\ln \mid \frac{1}{\sqrt{3}}\mid +\frac{1}{2})]$$$$I=4\ln 3-4$$

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