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
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