Problems on Techniques of Integration

Use the Integration by Parts. Set

$$\left\{\begin{array}{lll} u &=& x^2\\ dv &=& \sin(x) dx\;. \end{array}\right.$$

Then

$$\left\{\begin{array}{lll} du &=&2 x dx\\ v &=& -\cos(x)\;. \end{array}\right.$$

So

$$\int x^2 \sin(x) dx = - x^2 \cos(x) + \int 2 x \cos(x) dx = - x^2 \cos(x) + 2 \int x \cos(x) dx\;.$$

In order to integrate the function $x \cos(x)$, we will need to do another integration by parts. Set

$$\left\{\begin{array}{lll} u &=&x\\ dv &=& \cos(x)dx\;. \end{array}\right.$$

Then

$$\left\{\begin{array}{lll} du &=&dx\\ v &=& \sin(x)\;. \end{array}\right.$$

So

$$\int x \cos(x) dx = x \sin(x) - \int \sin(x) dx = x \sin(x) + \cos(x)\;.$$

Hence

$$\int x^2 \sin(x) dx = - x^2 \cos(x) + x \sin(x) + \cos(x) + C\;.$$

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