Mathematics

Find $\int {{{{{\sin }^6}x} \over {{{\cos }^8}x}}dx}$

SOLUTION
$\displaystyle \int{\dfrac{{\sin}^{6}{x}}{{\cos}^{8}{x}}dx}$
$=\displaystyle \int{{\tan}^{6}{x}{\sec}^{2}{x}dx}$
Let $t=\tan{x}\Rightarrow dt={\sec}^{2}{x}dx$
$=\displaystyle \int{{t}^{6}dt}$
$=\dfrac{{t}^{7}}{7}+c$ where $c$ is the constant of integration.
$=\dfrac{{\tan}^{7}{x}}{7}+c$ where $t=\tan{x}$

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

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