Mathematics

Evaluate the following integrals:
$$\int { \sin ^{ 3 }{ x } \cos ^{ 5 }{ x }  } dx$$


SOLUTION
$$\displaystyle\int{{\sin}^{3}{x}{\cos}^{5}{x}dx}$$

$$=\displaystyle\int{{\sin}^{3}{x}{\cos}^{2}{x}{\cos}^{3}{x}dx}$$

$$=\displaystyle\int{{\sin}^{3}{x}\left(1-{\sin}^{2}{x}\right){\cos}^{3}{x}dx}$$

$$=\displaystyle\int{\left({\sin}^{3}{x}-{\sin}^{5}{x}\right){\cos}^{3}{x}dx}$$

$$=\displaystyle\int{\left({\sin}^{3}{x}-{\sin}^{5}{x}\right){\cos}^{2}{x}\cos{x}dx}$$

$$=\displaystyle\int{\left({\sin}^{3}{x}-{\sin}^{5}{x}\right)\left(1-{\sin}^{2}{x}\right)\cos{x}dx}$$

$$=\displaystyle\int{\left({\sin}^{3}{x}-2{\sin}^{5}{x}+{\sin}^{7}{x}\right)\cos{x}dx}$$

$$=\displaystyle\int{{\sin}^{3}{x}\cos{x}dx}-2\displaystyle\int{{\sin}^{5}{x}\cos{x}dx}+\displaystyle\int{{\sin}^{7}{x}\cos{x}dx}$$

Let

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

$$=\displaystyle\int{{t}^{3}dt}-2\displaystyle\int{{t}^{5}dt}+\displaystyle\int{{t}^{7}dt}$$

$$=\dfrac{{t}^{4}}{4}-2\times\dfrac{{t}^{6}}{6}+\dfrac{{t}^{8}}{8}+c$$

$$=\dfrac{{\sin}^{4}{x}}{4}-\dfrac{{\sin}^{6}{x}}{3}+\dfrac{{\sin}^{8}{x}}{8}+c$$ where 

$$t=\sin{x}$$
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Subjective Medium Published on 17th 09, 2020
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