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Question Number 41174 by Tawa1 last updated on 02/Aug/18

Answered by MJS last updated on 03/Aug/18

$$\int {\mathrm{e}}^{x\sin x+\cos x}({\cos }^{2}x-\frac{\sin x}{x}-\frac{\cos x}{x^2})dx=$$$$\int {\mathrm{e}}^{x\sin x+\cos x}{\cos }^{2}xdx-\int \frac{{\mathrm{e}}^{x\sin x+\cos x}\sin x}{x}dx-\int \frac{{\mathrm{e}}^{x\sin x+\cos x}\cos x}{x^2}dx$$$$\mathrm{now}\mathrm{the}\mathrm{great}\mathrm{trick}:$$$$\int \frac{{\mathrm{e}}^{x\sin x+\cos x}\cos x}{x^2}dx=$$$$$$$$\int f^{\prime }g=fg-\int {fg}^{\prime }$$$$f^{\prime }=\frac{1}{x^2}\Rightarrow f=-\frac{1}{x}$$$$g={\mathrm{e}}^{x\sin x+\cos x}\cos x\Rightarrow g^{\prime }=x{\mathrm{e}}^{x\sin x+\cos x}{\cos }^{2}x-{\mathrm{e}}^{x\sin x+\cos x}\sin x$$$$$$$$=-\frac{1}{x}{\mathrm{e}}^{x\sin x+\cos x}\cos x-\int \frac{1}{x^2}(x{\mathrm{e}}^{x\sin x+\cos x}{\cos }^{2}x-{\mathrm{e}}^{x\sin x+\cos x}\sin x)dx$$$$$$$$\Rightarrow \mathrm{the}\mathrm{searched}\mathrm{integral}\mathrm{is}$$$$\frac{{\mathrm{e}}^{x\sin x+\cos x}\cos x}{x}+C$$

Commented by Tawa1 last updated on 03/Aug/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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