To solve:
∫ e sin ( 1 x ) 3 x 2 sec ( 1 x ) d x \int\dfrac{{\sqrt[3]{\mathrm{e}^{\sin{(\frac{1}{x})}}}}}{x^2\sec(\frac{1}{x})}dx ∫ x 2 sec ( x 1 ) 3 e s i n ( x 1 ) d x
We can re-write the above as:
∫ e sin ( 1 x ) 3 sec ( 1 x ) x 2 d x = ∫ cos ( 1 x ) e sin ( 1 x ) 3 x 2 d x \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 ( x 1 ) x 2 e 3 s i n ( x 1 ) d x = ∫ x 2 cos ( x 1 ) e 3 s i n ( x 1 ) d x
Substitute u = s i n ( 1 x ) 3 u = \dfrac{sin(\frac{1}{x})}{3} u = 3 s in ( x 1 )
∴ d u d x = c o s ( 1 x ) 3 x 2 ⟹ d x = − 3 x 2 c o s ( 1 x ) d u \therefore \quad \dfrac{du}{dx}= \dfrac{cos(\frac{1}{x})}{3x^2} \implies dx = -\dfrac{3x^2}{cos(\frac{1}{x})}du ∴ d x d u = 3 x 2 cos ( x 1 ) ⟹ d x = − cos ( x 1 ) 3 x 2 d u
Since u = s i n ( 1 x ) 3 u = \dfrac{sin(\frac{1}{x})}{3} u = 3 s in ( x 1 ) , then we have: − 3 ∫ e u d u = − 3 e u ⟹ − 3 e s i n ( 1 x ) 3 + C -3\int \mathrm{e}^u du = -3\mathrm{e}^u \implies -3\mathrm{e}^\frac{sin(\frac{1}{x})}{3}
+ C − 3 ∫ e u d u = − 3 e u ⟹ − 3 e 3 s in ( x 1 ) + C
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