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Question Number 152061 by mnjuly1970 last updated on 25/Aug/21

$$$$$$solve$$$$$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{{sin}^3\left(x\right).{cos}^2\left(x\right)}{x^3}dx\stackrel{?}{=}\frac{7\pi }{32}...◼$$

Answered by Ar Brandon last updated on 25/Aug/21

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{{\sin }^{3}x{\cos }^{2}x}{x^3}dx={\int }_0^{\mathrm{\infty }}\frac{{\sin }^{3}x-{\sin }^{5}x}{x^3}dx$$$$-8i{\sin }^{3}x={(e^{ix}-e^{-ix})}^3=e^{3ix}-3e^{ix}+3e^{-ix}-e^{3ix}$$$${\sin }^{3}x=\frac{3}{4}\sin x-\frac{1}{4}\sin 3x$$$$32i{\sin }^{5}x={(e^{ix}-e^{-ix})}^5=e^{5ix}-5e^{3ix}+10e^{ix}-10e^{-ix}+5e^{-3ix}-e^{-5ix}$$$${\sin }^{5}x=\frac{1}{16}\sin 5x-\frac{5}{16}\sin 3x+\frac{5}{8}\sin x$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}(\frac{1}{8}\cdot \frac{\sin x}{x^3}+\frac{1}{16}\cdot \frac{\sin 3x}{x^3}-\frac{1}{16}\cdot \frac{\sin 5x}{x^3})dx$$$$=\frac{1}{8}\cdot \frac{\pi }{2\mathrm{\Gamma }\left(3\right)\sin \left(\frac{3\pi }{2}\right)}+\frac{1}{16}\cdot \frac{\pi \times 3^2}{2\mathrm{\Gamma }\left(3\right)\sin \left(\frac{3\pi }{2}\right)}-\frac{1}{16}\cdot \frac{\pi \times 5^2}{2\mathrm{\Gamma }\left(3\right)\sin \left(\frac{3\pi }{2}\right)}$$$$=-\frac{\pi }{32}-\frac{9\pi }{64}+\frac{25\pi }{64}=\frac{-2\pi -9\pi +25\pi }{64}=\frac{14\pi }{64}=\frac{7\pi }{32}★$$

Commented by mnjuly1970 last updated on 25/Aug/21

$$thanksalotmasterbrandon$$

Commented by Ar Brandon last updated on 25/Aug/21

$${\mathrm{You}}^{\prime }\mathrm{re}\mathrm{welcome}\mathrm{Sir}!$$

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