Problems on Techniques of Integration

The trigonometric functions are nice functions since their derivatives and antiderivatives are easy to get. So for the integration by parts, these functions have the same behavior whether we differentiate them or take their antiderivatives. Therefore the focus should be on the other function involved in the integration. In this case, we must differentiate $f(x) = x^2$ because its derivative will lower its degree.

Set

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

Then

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

Since

$$\int u dv = u v - \int v du\;,$$

we get

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

In order to integrate the function $x \cos(x)$, we will need to do another integration by parts. This is very common that an integration by parts may lead to another one or may be more integration by parts. Set

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

Then

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

Since

$$\int u dv = u v - \int v du\;,$$

we get

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

Hence

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

It is a common mistake to forget the constant $C$.

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