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Question Number 142060 by iloveisrael last updated on 26/May/21

Answered by Ar Brandon last updated on 26/May/21

$$\mathrm{I}=\int \frac{1}{{\mathrm{x}}^2}{\mathrm{e}}^{\frac{1}{\mathrm{x}}}\sec (1+{\mathrm{e}}^{\frac{1}{\mathrm{x}}})\tan (1+{\mathrm{e}}^{\frac{1}{\mathrm{x}}})\mathrm{dx}$$$$u=1+{\mathrm{e}}^{\frac{1}{\mathrm{x}}}\Rightarrow \mathrm{d}u=-\frac{1}{{\mathrm{x}}^2}{\mathrm{e}}^{\frac{1}{\mathrm{x}}}\mathrm{dx}$$$$\mathrm{I}=-\int \sec \left(u\right)\tan \left(u\right)du=-\sec \left(u\right)+\mathrm{C}$$$$=-\sec (1+{\mathrm{e}}^{\frac{1}{\mathrm{x}}})+\mathrm{C}$$

Answered by EDWIN88 last updated on 26/May/21

$$\mathrm{J}=\int \frac{{\mathrm{e}}^{\frac{1}{\mathrm{x}}}}{{\mathrm{x}}^2}\frac{\sin (1+{\mathrm{e}}^{\frac{1}{\mathrm{x}}})}{\cos {}^2(1+{\mathrm{e}}^{\frac{1}{\mathrm{x}}})}\mathrm{dx}$$$$\mathrm{let}\mathrm{y}=\cos (1+{\mathrm{e}}^{\frac{1}{\mathrm{x}}})\Rightarrow \mathrm{dy}=\frac{{\mathrm{e}}^{\frac{1}{\mathrm{x}}}}{{\mathrm{x}}^2}.\sin (1+{\mathrm{e}}^{\frac{1}{\mathrm{x}}})\mathrm{dx}$$$$\mathrm{J}=\int \frac{\mathrm{dy}}{{\mathrm{y}}^2}=-\frac{1}{\mathrm{y}}+\mathrm{c}$$$$\mathrm{J}=-\sec (1+{\mathrm{e}}^{\frac{1}{\mathrm{x}}})+\mathrm{c}$$

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