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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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