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Question Number 17676 by mondodotto@gmail.com last updated on 09/Jul/17

Answered by alex041103 last updated on 09/Jul/17

$$\mathrm{First}:$$$$\frac{{5\cos }^3\mathrm{x}+{2\sin }^3\mathrm{x}}{{2\sin }^2{\mathrm{xcos}}^2\mathrm{x}}=\frac{5}{2}\frac{\mathrm{cosx}}{{\sin }^2\mathrm{x}}+\frac{\mathrm{sinx}}{{\cos }^2\mathrm{x}}$$$$\Rightarrow \mathrm{I}=\int \frac{{5\cos }^3\mathrm{x}+{2\sin }^3\mathrm{x}}{{2\sin }^2{\mathrm{xcos}}^2\mathrm{x}}=$$$$=\frac{5}{2}\int \frac{\mathrm{cosx}}{{\sin }^2\mathrm{x}}\mathrm{dx}+\int \frac{\mathrm{sinx}}{{\cos }^2\mathrm{x}}\mathrm{dx}$$$$\mathrm{No}\mathrm{We}\mathrm{substitute}\mathrm{for}\mathrm{the}\mathrm{first}\mathrm{integral}$$$$\mathrm{u}=\mathrm{sinx}\to \mathrm{du}=\mathrm{cosxdx}$$$$\mathrm{And}\mathrm{for}\mathrm{the}\mathrm{second}$$$$\mathrm{w}=\mathrm{cosx}\to \mathrm{dw}=-\mathrm{sinxdx}$$$$\Rightarrow \mathrm{I}=\frac{5}{2}\int {\mathrm{u}}^{-2}\mathrm{du}-\int {\mathrm{w}}^{-2}\mathrm{dw}=$$$$=-\frac{5}{2\mathrm{sinx}}+\frac{1}{\mathrm{cosx}}+\mathrm{C}$$$$\Rightarrow \mathrm{I}=\int \frac{{5\cos }^3\mathrm{x}+{2\sin }^3\mathrm{x}}{{2\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}=\frac{1}{\mathrm{cosx}}-\frac{5}{2\mathrm{sinx}}+\mathrm{C}$$

Commented by mondodotto@gmail.com last updated on 09/Jul/17

$$\mathrm{thanx}\mathrm{really}\mathrm{appreciated}\mathrm{sir}$$

Answered by Arnab Maiti last updated on 11/Jul/17

$$=\int \frac{5}{2}\frac{\mathrm{cosx}}{{\sin }^2\mathrm{x}}\mathrm{dx}+\int \frac{\mathrm{sinx}}{{\cos }^2\mathrm{x}}\mathrm{dx}$$$$=\frac{5}{2}\int \mathrm{cosecx}\mathrm{cotx}\mathrm{dx}+\int \mathrm{tanx}\mathrm{secx}\mathrm{dx}$$$$=-\frac{5}{2}\mathrm{cosecx}+\mathrm{secx}+\mathrm{C}$$

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