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Question Number 114689 by arkanmath7@gmail.com last updated on 20/Sep/20

$$\int {xsin}^{n}xdx$$

Answered by 1549442205PVT last updated on 20/Sep/20

$$\mathrm{Just}\mathrm{give}\mathrm{out}\mathrm{reduction}\mathrm{formula}:$$$$\mathrm{Integrating}\mathrm{by}\mathrm{parts}\mathrm{we}\mathrm{get}:$$$${\mathrm{I}}_{\mathrm{n}}=\int {xsin}^{n}xdx=\int {\mathrm{xsin}}^{\mathrm{n}-1}\mathrm{x}.\mathrm{sinxdx}$$$$=-{\mathrm{xsin}}^{\mathrm{n}-1}\left(\mathrm{x}\right)\mathrm{cosx}+\int \mathrm{cosx}\{{\sin }^{\mathrm{n}-1}\mathrm{x}+\mathrm{x}(\mathrm{n}-1){\sin }^{\mathrm{n}-2}\mathrm{xcosx}\}\mathrm{dx}$$$$=-{\mathrm{xcosxsin}}^{\mathrm{n}-1}\mathrm{x}+\int {\sin }^{\mathrm{n}-1}\mathrm{xcosxdx}$$$$+(\mathrm{n}-1)\int {\mathrm{xsin}}^{\mathrm{n}-2}\mathrm{x}(1-{\sin }^2\mathrm{x})\mathrm{dx}$$$$=-{\mathrm{xcosxsin}}^{\mathrm{n}-1}\mathrm{x}+\frac{{\sin }^{\mathrm{n}}\mathrm{x}}{\mathrm{n}}+(\mathrm{n}-1)({\mathrm{I}}_{\mathrm{n}-2}-{\mathrm{I}}_{\mathrm{n}})$$$$\Rightarrow (1+(\mathrm{n}-1){\mathrm{I}}_{\mathrm{n}}=-{\mathrm{xcosxsin}}^{\mathrm{n}-1}\mathrm{x}+\frac{{\sin }^{\mathrm{n}}\mathrm{x}}{\mathrm{n}}$$$$+(\mathrm{n}-1){\mathrm{I}}_{\mathrm{n}-2}$$$$\Rightarrow {\mathrm{I}}_{\mathrm{n}}=\frac{-{\mathrm{xcosxsin}}^{\mathrm{n}-1}\mathrm{x}}{\mathrm{n}}+\frac{{\sin }^{\mathrm{n}}\mathrm{x}}{{\mathrm{n}}^2}+\frac{\mathrm{n}-1}{\mathrm{n}}{\mathrm{I}}_{\mathrm{n}-2}$$

Commented by arkanmath7@gmail.com last updated on 20/Sep/20

$$IsolveditinsamewaybutIstoppedat$$$$I_{n-2}.Isittruetoletitasitis?$$

Commented by 1549442205PVT last updated on 20/Sep/20

$$\mathrm{Just}\mathrm{need}\mathrm{above}\mathrm{formula}\mathrm{we}\mathrm{will}\mathrm{find}$$$$\mathrm{out}{\mathrm{I}}_{\mathrm{n}}\mathrm{for}\mathrm{\forall }\mathrm{n}\in {\mathrm{N}}^{\ast }$$$${\mathrm{I}}_{\mathrm{n}}\Leftarrow {\mathrm{I}}_{\mathrm{n}-2}\Leftarrow {\mathrm{I}}_{\mathrm{n}-4}\Leftarrow ....\Leftarrow {\mathrm{I}}_1\left(\mathrm{if}\mathrm{n}\mathrm{odd}\right)$$$${\mathrm{I}}_{\mathrm{n}}\Leftarrow {\mathrm{I}}_{\mathrm{n}-2}\Leftarrow ....\Leftarrow {\mathrm{I}}_2\left(\mathrm{if}\mathrm{n}\mathrm{even}\right)$$$${\mathrm{I}}_1,{\mathrm{I}}_2\mathrm{are}\mathrm{simple}\mathrm{cases}$$

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