Mathematics

Evaluate:
$$\displaystyle \int_{-1}^{1}\sin^{5}x\cos^{4}x\ dx$$


SOLUTION
$$\int_{-1}^{1}{\sin^{5}{x} \cos^{4}{x} \; dx}$$
$$= \int_{-1}^{1}{\sin^{4}{x} \cos^{4}{x} \; \sin{x} \; dx}$$
$$= \int_{-1}^{1}{{\left( 1 - \cos^{2}{x} \right)}^{2} \cos^{4}{x} \sin{x} \; dx}$$
$$= \int_{-1}^{1}{\left( 1 + \cos^{4}{x} - 2 \cos^{2}{x} \right) \cos^{4}{x} \sin{x} \; dx}$$
$$= \int_{-1}^{1}{\left( \cos^{4}{x} + \cos^{8}{x} - 2 \cos^{6}{x} \right) \sin{x} \; dx}$$
Let $$\cos{x} = t \Rightarrow - \sin{x} \; dx = dt$$
Also,
At $$x = -1 \Rightarrow t = \cos{-1} = \cos{\left( 1 \right)}$$
At $$x = 1 \Rightarrow t = \cos{\left( 1 \right)}$$
Therefore,
$$\int_{\cos{\left( 1 \right)}}^{\cos{\left( 1 \right)}}{\left( {t}^{4} + {t}^{8} - 2 {t}^{6} \right) dt}$$
$$= 0$$
Thus,
$$\int_{-1}^{1}{\sin^{5}{x} \cos^{4}{x} \; dx} = 0$$
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Subjective Medium Published on 17th 09, 2020
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