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
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