Mathematics

Solve $\displaystyle \int{\sin^{3}x\cos^{5}xdx}$.

SOLUTION
Let $I = \int{\sin^{3}{x} \cos^{5}{x} dx}$

$= \int{\sin^{2}{x} \cdot \sin{x} \cos^{5}{x} dx}$

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

$= \int{\cos^{5}{x} \sin{x} dx} - \int{\cos^{7}{x} \sin{x} dx}$

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

Therefore,
$I = \int{{t}^{7} dt} - \int{{t}^{5} dt}$

$\Rightarrow I = \cfrac{{t}^{8}}{8} - \cfrac{{t}^{6}}{6} + C$

$\Rightarrow I = \cfrac{\cos^{8}}{8} - \cfrac{\cos^{6}}{6} + C$

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

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