Mathematics

Evaluate the following integral:
$$\int { \sqrt [ 3 ]{ \cos ^{ 2 }{ x }  } \sin { x }  } dx\quad $$


SOLUTION
Let 
$$t=\cos{x}\Rightarrow\,dt=-\sin{x}dx$$

$$\Rightarrow\,-dt=\sin{x}dx$$

$$\displaystyle\int{\sqrt[3]{{\cos}^{2}{x}}\sin{x}\,dx}$$

$$=-\displaystyle\int{\sqrt[3]{{t}^{2}}dt}$$

$$=-\displaystyle\int{{t}^{\frac{2}{3}}dt}$$

$$=-\dfrac{{t}^{\frac{2}{3}+1}}{\dfrac{2}{3}+1}+c$$

$$=-\dfrac{{t}^{\frac{5}{3}}}{\dfrac{5}{3}}+c$$

$$=-\dfrac{3}{5}{t}^{\frac{5}{3}}+c$$

$$=-\dfrac{3}{5}{\cos}^{\frac{5}{3}}{x}+c$$ where $$t=\cos{x}$$

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Subjective Medium Published on 17th 09, 2020
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