a lecture on integration by parts

The Method of Integration by Parts

This method is used to find indefinite integrals that look like the result of a product rule.

• When to use it: Use this method when the integrand is a product of two functions (and when the method ofsubstitution clearly does not work).

• How to use it: The method is based in the following formula:

  $$\int U\cdot V^{\prime }𝑑x=U\cdot V-\int U^{\prime }\cdot V𝑑x$$

  Suppose we want to solve $\int f(x)𝑑x$:

  1. Find functions $U$ and $V^{\prime }$ such that $U\cdot V^{\prime }=f(x)$. There are many possible choices.

  2. $U$ should be a function easy to derive and such that the derivative of $U$ is easier, less complicatedthan $U$ itself. For example good choices for $U$ are $U=x,x^2,x^3,e^x,\ln x$.

  3. $V^{\prime }$ should be a function which is easy to integrate, such that we can find $\int V^{\prime }𝑑x$ easily and the integralis less complicated than $V^{\prime }$ itself. Good choicesfor $V^{\prime }$ are $V^{\prime }=e^x,\sin x,\cos x$. The functions $x,x^2,x^3,\ln x$ are bad choices.

  4. Once the functions $U$ and $V^{\prime }$ are chosen, find $U^{\prime }=\frac{d}{dx}U$ and $V=\int V^{\prime }𝑑x$.

  5. Plug in the formula.

  6. Solve the new integral $\int U^{\prime }\cdot V𝑑x$, which if the choices of $U$ and $V^{\prime }$ were good, should be easy.

  7. If the new integral is hard, the choices of $U$ and $V^{\prime }$ might be wrong. So repeat the choice.

Again, the method is best explained through examples:

Example 0.1.

Find $\int x{e}^x𝑑x$. This is a product of two functions and there is no composition visible, so we will be usingintegration by parts. We need two functions $U$ and $V^{\prime }$ such that $U\cdot V^{\prime }=x{e}^x$. A good choice for $U$ is $x$ because$U^{\prime }={(x)}^{\prime }=1$ and the derivative is simpler than $x$ itself. A good choice for $V^{\prime }=e^x$ because it is easy to integrate,$\int e^x𝑑x=e^x$ (and $U\cdot V^{\prime }=x{e}^x$). Therefore we apply the formula:

$$\int x{e}^x𝑑x=UV-\int U^{\prime }V𝑑x=x{e}^x-\int 1\cdot e^x𝑑x=x{e}^x-\int e^x𝑑x=x{e}^x-e^x+C.$$

Example 0.2.

Find:

$$\int x\cos (x)𝑑x.$$

We choose $U=x$ and $V^{\prime }=\cos (x)$. Then $U^{\prime }=1$ and $V=\int V^{\prime }𝑑x=\sin (x)$. Apply the formula:

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

Example 0.3.

Find $\int \ln xdx$. Although this is not the product of two functions, $\ln x$ is a typical example of a function to beintegrated by parts. Here is how, let $U=\ln x$ and $V^{\prime }=1$ so that $U^{\prime }=1/x$ and $V=x$. Look what happens when we use the formula:

$$\int \ln xdx=x\ln x-\int \frac{1}{x}\cdot x𝑑x=x\ln x-\int 1𝑑x=x\ln x-x+C.$$

Example 0.4.

In some cases we have to use the method of integration by parts twice to get an answer. For example find $\int x^2\sin xdx$.We put $U=x^2$ and $V^{\prime }=\sin x$, and so $U^{\prime }=2x$ and $V=\int \sin xdx=-\cos x$. Thus:

$$\int x^2\sin xdx=-x^2\cos x+\int 2x\cos xdx=-x^2\cos x+2\int x\cos xdx$$

and above, we have seen that in order to find $\int x\cos xdx$ we use integration by parts. Therefore the final anwser is:

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

Example 0.5.

Finally, in some other cases, after we do integration by parts twice, we get to the same integral we wanted to solve.Although it would seem we are stuck, no no! we will be able to find a solution right away.

Find $\int e^x\sin xdx$. Let $I=\int e^x\sin xdx$. We start with integration by parts, taking

$$U=e^x,V^{\prime }=\sin x,U^{\prime }=e^x,V=-\cos x$$

Thus:

$$I=-e^x\cos x+\int e^x\cos xdx$$

Hmmm…the integral $\int e^x\cos xdx$ looks like the one we started with…we use integration by parts to solve this one.Take $U=e^x$ and $V^{\prime }=\cos x$ so $U^{\prime }=e^x$ and $V=\sin x$. Thus:

$$\int e^x\cos xdx=e^x\sin x-\int e^x\sin xdx.$$

Therefore:

$$I=-e^x\cos x+(e^x\sin x-\int e^x\sin xdx)=-e^x\cos x+e^x\sin x-I$$

Remember that we want to find $I$, so if we solve for $I$ above, we obtain:

$$2I=-e^x\cos x+e^x\sin x$$

and so $I=1/2(-e^x\cos x+e^x\sin x)+C$.