Integration by parts

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Figure 5.4 shows a technique called integration by parts. If the integral $\int vdu$ is easier than the integral $\int udv$ , then we can calculate the easier one, and then by simple geometry determine the one we wanted. Identifying the large rectangle that surrounds both shaded areas, and the small white rectangle on the lower left, we have

$\int udv=(area of large rectangle)-(area of small rectangle)\int vdu$

In the case of an indefinite integral, we have a similar relationship de- rived from the product rule:

$\begin{align*} d(uv) &=udv+vdu \\ udv&=d(uv)-vdu \end{align*}$

Integrating both sides, we have the following relation.

Integration by parts

$\int udv=uv-\int vdu$

Figure 5.4 Integration by parts

Since a definite integral can always be done by evaluating an indefinite integral at its upper and lower limits, one usually uses this form. Integrals don't usually come prepackaged in a form that makes it obvious that you should use integration by parts. What the equation for integration by parts tells us is that if we can split up the integrand into two factors, one of which (the ) we know how to integrate, we have the option of changing the integral into a new form in which that factor becomes its integral, and the other factor becomes its derivative. If we choose the right way of splitting up the integrand into parts, the result can be a simplification.

Example

Evaluate

$\int x\textrm{ cos }x dx$

There are two obvious possibilities for splitting up the integrand into factors,

$udv=(x)(\textrm{cos }x dx)$ or

$udv=(\textrm{cos }x)(xdx)$

The first one is the one that lets us make progress. If u=x , then du=dx , and if $dv=\textrm{cos }x dx$ , then integration gives $v=\textrm{sin }x$ .

$\begin{align*} \int x\textrm{ cos }x dx &=\int u dv \\ &= uv-\int vdu\\ &=x\textrm{ sin }x-\int \textrm{sin }x dx \\ &= x\textrm{ sin }x + \textrm{cos }x \end{align*}$

Of the two possibilities we considered for and , the reason this one helped was that differentiating gave , which was simpler, and integrating $\textrm{cos }x dx$ gave $\textrm{sin }x$ , which was no more complicated than before. The second possibility would have made things worse rather than better, because integrating xdx would have given x^2/2 , which would have been more complicated rather than less.

Example

Evaluate $\int \textrm{In }x dx$

This one is a little tricky, because it isn’t explicitly written as a product, and yet we can attack it using integration by parts. Let $u=\textrm{In }x$ and dv=dx
$\begin{align*} \int \textrm{In }x dx &=\int udv \\ &= uv-\int v du\\ &= x \textrm{In} x-\int x\frac{dx}{x}\\ &= x \textrm{In} x -x \end{align*}$

Example

Evaluate $\int x^2e^xdx$

Integration by parts lets us split the integrand into two factors, integrate one, differentiate the other, and then do that integral. Integrating or differentiating e^x does nothing. Integrating x^2 increases the exponent, which makes the problem look harder, whereas differentiating x^2 knocks the exponent down a step, which makes it look easier. Let u=x^2 and dv=e^xdx , so that du=2xdx and v=e^x . We then have

$\int x^2e^xdx=x^2e^x-2\int xe^xdx$ Although we don’t immediately know how to evaluate this new integral, we can subject it to the same type of inte- gration by parts, now with u=x and dv=e^xdx . After the second integra- tion by parts, we have: