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Question Number 149415 by abdurehime last updated on 05/Aug/21

$$\mathrm{show}\mathrm{that}{\int }_0^{\mathrm{\infty }}{\left(\frac{\mathrm{sint}-\mathrm{tcost}}{{\mathrm{t}}^3}\right)}^2\mathrm{dt}=\frac{\mathrm{\Pi }}{15}$$

Commented by abdurehime last updated on 05/Aug/21

$$\mathrm{please}\mathrm{help}\mathrm{me}????$$

Answered by mindispower last updated on 05/Aug/21

$$byart$$$$=[-\frac{1}{5}\frac{{(sin\left(t\right)-tcos\left(t\right))}^2}{t^5}]$$$$+\frac{2}{5}{\int }_0^{\mathrm{\infty }}\frac{(sin\left(t\right)-tcos\left(t\right))sin\left(t\right)}{t^4}dt$$$$=\frac{2}{5}{\left[\frac{(sin\left(t\right)-tcos\left(t\right))sin\left(t\right)}{-3t^3}\right]}_0^{\mathrm{\infty }}+\frac{2}{15}{\int }_0^{\mathrm{\infty }}\frac{(sin\left(t\right)cos\left(t\right)-tcos\left(2t\right))}{t^3}dt$$$$=\frac{1}{15}{\left[\frac{sin\left(t\right)cos\left(t\right)-tcos\left(2t\right)}{-t^2}\right]}_0^{\mathrm{\infty }}+\frac{1}{15}{\int }_0^{\mathrm{\infty }}\frac{2tsin\left(2t\right)}{t^2}$$$$=\frac{1}{15}{\int }_0^{\mathrm{\infty }}\frac{sin\left(2t\right).2dt}{t}=\frac{2}{15}{\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)dx}{x}=\frac{2}{15}.\frac{\pi }{2}=\frac{\pi }{15}$$$$$$

Commented by abdurehime last updated on 05/Aug/21

$$\mathrm{please}\mathrm{show}\mathrm{me}\mathrm{step}\mathrm{by}\mathrm{step}$$

Commented by mnjuly1970 last updated on 05/Aug/21

$$verynicesirpower...beavo..$$

Commented by mindispower last updated on 05/Aug/21

$$thanxsir$$

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