Mathematics

Evaluate the given integral.$\int { { x }^{ 2 }\cos { x } } dx$

SOLUTION
$I=\displaystyle\int{{x}^{2}\cos{x}dx}$

Integrating by parts

Let $u={x}^{2}\Rightarrow\,du=2x\,dx$

$dv=\cos{x}dx\Rightarrow\,v=\sin{x}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$......by parts formula.

$\Rightarrow\,I={x}^{2}\sin{x}-\displaystyle\int{2x\sin{x}\,dx}$

$\Rightarrow\,I={x}^{2}\sin{x}-2\displaystyle\int{x\sin{x}\,dx}$

Let $u=x\Rightarrow\,du=dx$ and $dv=\sin{x}\,dx\Rightarrow\,v=-\cos{x}$

$\Rightarrow\,I={x}^{2}\sin{x}-2\left[-x\cos{x}+\displaystyle\int{\cos{x}dx}\right]$

$\Rightarrow\,I={x}^{2}\sin{x}-2\left[-x\cos{x}+\sin{x}\right]+c$

$\therefore\,I={x}^{2}\sin{x}+2x\cos{x}-2\sin{x}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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