Integrate (cbrte^sin(1/x))/(x^2 sec(1/x))
To solve:
∫esin(1x)3x2sec(1x)dx\int\dfrac{{\sqrt[3]{\mathrm{e}^{\sin{(\frac{1}{x})}}}}}{x^2\sec(\frac{1}{x})}dx∫x2sec(x1)3esin(x1)dx
We can re-write the above as:
∫esin(1x)3sec(1x)x2dx=∫cos(1x)esin(1x)3x2dx\int \dfrac{\mathrm{e}^\frac{\sin\left(\frac{1}{x}\right)}{3}}{\sec\left(\frac{1}{x}\right)x^2} dx \quad = \int \dfrac{\cos(\frac{1}{x})\mathrm{e}^\frac{\sin\left(\frac{1}{x}\right)}{3}}{x^2} dx∫sec(x1)x2e3sin(x1)dx=∫x2cos(x1)e3sin(x1)dx
Substitute u=sin(1x)3u = \dfrac{sin(\frac{1}{x})}{3}u=3sin(x1)
∴dudx=cos(1x)3x2 ⟹ dx=−3x2cos(1x)du\therefore \quad \dfrac{du}{dx}= \dfrac{cos(\frac{1}{x})}{3x^2} \implies dx = -\dfrac{3x^2}{cos(\frac{1}{x})}du∴dxdu=3x2cos(x1)⟹dx=−cos(x1)3x2du
Since u=sin(1x)3u = \dfrac{sin(\frac{1}{x})}{3}u=3sin(x1) , then we have: −3∫eudu=−3eu ⟹ −3esin(1x)3+C-3\int \mathrm{e}^u du = -3\mathrm{e}^u \implies -3\mathrm{e}^\frac{sin(\frac{1}{x})}{3} + C−3∫eudu=−3eu⟹−3e3sin(x1)+C
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