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Question Number 50466 by afachri last updated on 16/Dec/18

$$$$$$\int \frac{\cos \left(2t\right)}{t^3}dt=....$$

Commented by afachri last updated on 16/Dec/18

$$\mathbf{Would}\mathbf{anybody}\mathbf{give}\mathbf{me}\mathbf{a}\mathbf{hand}$$$$\mathbf{on}\mathbf{this}?{\mathbf{i}}^{\prime }\mathbf{m}\mathbf{stuck}.$$

Commented by maxmathsup by imad last updated on 16/Dec/18

$$atformofseriewehavecos\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{2n}}{\left(2n\right)!}\Rightarrow$$$$cos\left(2t\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\left(2n\right)!}{2}^{2n}{t}^{2n}\Rightarrow \int \frac{cos\left(2t\right)}{t^3}dt$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{4}^n}{\left(2n\right)!}\int t^{2n-3}dt=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{4^n}{\left(2n\right)!(2n-2)}{t}^{2n-2}+c.$$$$$$

Commented by afachri last updated on 17/Dec/18

$$\mathrm{aha},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{McLaurin}\mathrm{series}\mathrm{Sir}.\mathrm{Thank}\mathrm{u}$$$$\mathrm{very}\mathrm{much}\mathrm{Sir}$$

Commented by maxmathsup by imad last updated on 17/Dec/18

$$youarewelcome.$$

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