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Question Number 124438 by Ar Brandon last updated on 03/Dec/20

$$\int {\mathrm{e}}^{(\mathrm{xsinx}+\mathrm{cosx})}\cdot \left(\frac{{\mathrm{x}}^4{\cos }^3\mathrm{x}-\mathrm{xsinx}+\mathrm{cosx}}{{\mathrm{x}}^2{\cos }^2\mathrm{x}}\right)\mathrm{dx}$$

Commented by Dwaipayan Shikari last updated on 03/Dec/20

$$IIT-JEEbook?$$

Commented by Ar Brandon last updated on 03/Dec/20

Yeah ��������

Answered by Ar Brandon last updated on 03/Dec/20

$$\mathrm{\Omega }=\int {\mathrm{e}}^{(\mathrm{xsinx}+\mathrm{cosx})}\cdot \left(\frac{{\mathrm{x}}^4{\cos }^3\mathrm{x}-\mathrm{xsinx}+\mathrm{cosx}}{{\mathrm{x}}^2{\cos }^2\mathrm{x}}\right)\mathrm{dx}$$$$=\int {\mathrm{xe}}^{(\mathrm{xsinx}+\mathrm{cosx})}\cdot \mathrm{x}\cdot \mathrm{cosxdx}+\int {\mathrm{e}}^{(\mathrm{xsinx}+\mathrm{cosx})}\cdot \frac{-\mathrm{xsinx}+\mathrm{cosx}}{{\left(\mathrm{xcosx}\right)}^2}\mathrm{dx}$$$$=\{{\mathrm{xe}}^{(\mathrm{xsinx}+\mathrm{cosx})}-\int {\mathrm{e}}^{(\mathrm{xsinx}+\mathrm{cosx})}\mathrm{dx}\}+\{{\mathrm{e}}^{(\mathrm{xsinx}+\mathrm{cosx})}\cdot \frac{-1}{\mathrm{xcosx}}+\int {\mathrm{e}}^{(\mathrm{xsinx}+\mathrm{cosx})}\mathrm{dx}\}$$$$=(\mathrm{x}-\frac{1}{\mathrm{xcosx}}){\mathrm{e}}^{(\mathrm{xsinx}+\mathrm{cosx})}+\mathcal{C}$$

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