Mathematics

# solve it.$I = \int {{x^2}\cos x} dx$

##### SOLUTION
$I=\displaystyle\int{{x}^{2}\cos{x}dx}$
Let $u={x}^{2}\Rightarrow\,du=2x\,dx$
$dv=\cos{x}dx\Rightarrow\,v=\sin{x}$
$I={x}^{2}\sin{x}-\displaystyle\int{2x\sin{x}\,dx}$
$I={x}^{2}\sin{x}-2\displaystyle\int{x\sin{x}\,dx}$
Let $u=x\Rightarrow\,du=dx$
$dv=\sin{x}dx\Rightarrow\,v=-\cos{x}$
$I={x}^{2}\sin{x}-2\left[-x\cos{x}+\displaystyle\int{\cos{x}dx}\right]$
$I={x}^{2}\sin{x}+2x\cos{x}-2\displaystyle\int{\cos{x}dx}$
$I={x}^{2}\sin{x}+2x\cos{x}-2\sin{x}+c$

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

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