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Question Number 134811 by bramlexs22 last updated on 07/Mar/21

$${\int }_0^{\pi /2}\frac{\sin \left(\frac{3\mathrm{x}}{2}\right)}{\tan \left(3\mathrm{x}\right)}\mathrm{dx}$$

Answered by EDWIN88 last updated on 07/Mar/21

$$\mathrm{set}\frac{3\mathrm{x}}{2}=\mathrm{t}\Rightarrow 3\mathrm{x}=2\mathrm{t},\overline{)\begin{array}{c}\mathrm{x}=\frac{\pi }{2}\to \mathrm{t}=\frac{3\pi }{4}\\ \mathrm{x}=0\to \mathrm{t}=0\end{array}}$$$$\mathbb{L}={\int }_0^{3\pi /4}\frac{\sin \mathrm{t}}{\tan 2\mathrm{t}}\left(\frac{2}{3}\mathrm{dt}\right)$$$$\mathbb{L}=\frac{2}{3}{\int }_0^{3\pi /4}\frac{\cos 2\mathrm{t}\sin \mathrm{t}}{2\sin \mathrm{tcos}\mathrm{t}}\mathrm{dt}$$$$\mathbb{L}=\frac{1}{3}{\int }_0^{3\pi /4}\frac{\cos 2t}{\cos t}dt=\frac{1}{3}{\int }_0^{3\pi /4}\frac{2\cos {}^2\mathrm{t}-1}{\cos \mathrm{t}}\mathrm{dt}$$$$\mathbb{L}=\frac{1}{3}{[2\sin \mathrm{t}-\ln \mid \sec \mathrm{t}+\tan \mathrm{t}\mid ]}_0^{\frac{3\pi }{4}}$$$$\mathbb{L}=\frac{1}{3}[\sqrt{2}-\ln \mid \sqrt{2}+1\mid ]$$

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