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
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