Mathematics

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

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

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

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

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

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

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

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

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

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

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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